求数列 1×1!，2×2!，3×3!，4×4!，…，n×n! 前 n 项之和 Sn 的通项公式

费尔马1

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费尔马1

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王守恩

谢谢费尔马1! 挺不错的题目！谢谢费尔马1！新年快乐！！！
求数列 1×1!，2×2!，3×3!，4×4!，…，n×n! 前 n 项之和 Sn 的通项公式

A(n)=1, 5, 23, 119, 719, 5039, 40319, 362879, 3628799, 39916799, 479001599, 6227020799,
87178291199, 1307674367999, 20922789887999, 355687428095999, 6402373705727999,
121645100408831999, 2432902008176639999, 51090942171709439999,
1124000727777607679999, 25852016738884976639999, 620448401733239439359999,
15511210043330985983999999, 403291461126605635583999999,
10888869450418352160767999999, 304888344611713860501503999999,
8841761993739701954543615999999, 265252859812191058636308479999999,
8222838654177922817725562879999999,
263130836933693530167218012159999999,
8683317618811886495518194401279999999,
295232799039604140847618609643519999999,
10333147966386144929666651337523199999999,
371993326789901217467999448150835199999999,
13763753091226345046315979581580902399999999,
523022617466601111760007224100074291199999999,
......
B(n)=11, 71, 503, 4031, 36287, 36287, 399167, 4790015, 62270207,871782911, 1307674367,
20922789887, 355687428095, 6402373705727, 121645100408831, 243290200817663,
5109094217170943, 112400072777760767, 2585201673888497663, 62044840173323943935,
15511210043330985983, 403291461126605635583,
10888869450418352160767, 304888344611713860501503,
8841761993739701954543615, 26525285981219105863630847,
822283865417792281772556287,
26313083693369353016721801215,
868331761881188649551819440127,
29523279903960414084761860964351,
103331479663861449296666513375231,
3719933267899012174679994481508351,
137637530912263450463159795815809023,
5230226174666011117600072241000742911,
......
B(n)的规律是这样: 去掉A(1),A(2),A(3)；去掉A(4)末尾的9,去掉A(5)末尾的9,去掉A(6)末尾的9,
去掉A(7)末尾的9,去掉A(8)末尾的9,去掉A(9)末尾的9,......末尾连在一起的9都去掉, 具体见上。
B(n)的通项公式还是蛮有意思的(OEIS还是没有)。你搞不了，我也搞不了。
多年的老朋友 ！千言万语并一句: 新年快乐！！！

Treenewbee

\[A_n=(n+1)!-1\]

Treenewbee

王守恩 发表于 2024-1-14 19:54
谢谢费尔马1! 挺不错的题目！谢谢费尔马1！新年快乐！！！
求数列 1×1!，2×2!，3×3!，4×4!，…，n×n! ...

1. Array[(# + 4)!/10^IntegerExponent[(# + 4)!] - 1 &, 34]

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1. Array[(#+4)!//.x_/;x~Mod~5==0:>x/10-1&,34]

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{11,71,503,4031,36287,36287,399167,4790015,62270207,871782911,1307674367,20922789887,355687428095,6402373705727,121645100408831,243290200817663,5109094217170943,112400072777760767,2585201673888497663,62044840173323943935,15511210043330985983,403291461126605635583,10888869450418352160767,304888344611713860501503,8841761993739701954543615,26525285981219105863630847,822283865417792281772556287,26313083693369353016721801215,868331761881188649551819440127,29523279903960414084761860964351,103331479663861449296666513375231,3719933267899012174679994481508351,137637530912263450463159795815809023,5230226174666011117600072241000742911}

王守恩

1. Table[n!/10^Sum[Floor[n/5^i],{i,n}]-1,{n,5,38}]

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谢谢费尔马1！没主帖的数字串垫底, 光是凭上面那些数, 规律还真不好找。谢谢费尔马1！