A000540 -id:A000540 - cp's OEIS frontend

A123095 Sum of first n 11th powers.

0, 1, 2049, 179196, 4373500, 53201625, 415998681, 2393325424, 10983260016, 42364319625, 142364319625, 427675990236, 1170684360924, 2962844754961, 7012409924625, 15662165784000, 33254351828416, 67526248136049, 131794658215281, 248284917113500, 453084917113500

Offset: 0

Zerinvary Lajos, Sep 27 2006

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), A023002 (m=10), this sequence (m=11), A123094 (m=12), A181134 (m=13).
Cf. A008455, A215083.
Cf. A008275, A008277.

[(&+[j^11: j in [0..n]]): n in [0..30]]; // G. C. Greubel, Jul 21 2021

[seq(add(i^11, i=1..n), n=0..20)];
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^11 od: seq(a[n], n=0..13); # Zerinvary Lajos, Feb 22 2008

Table[Sum[k^11, {k, n}], {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
Accumulate[Range[0,20]^11] (* Harvey P. Dale, Sep 17 2021 *)

A123095_list, m = [0], [39916800, -199584000, 419126400, -479001600, 322494480, -129230640, 29607600, -3498000, 171006, -2046, 1, 0 , 0]
for _ in range(10**2):
    for i in range(12):
        m[i+1]+= m[i]
    A123095_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

[(bernoulli_polynomial(n+1, 12) - bernoulli(12))/12  for n in (0..30)] # G. C. Greubel, Jul 21 2021

a(n) = n*A023002(n) - Sum_{i=0..n-1} A023002(i). - Bruno Berselli, Apr 27 2010
a(n) = n^2*(n+1)^2*(2*n^8 +8*n^7 +4*n^6 -16*n^5 -5*n^4 +26*n^3 -3*n^2 -20*n +10)/24. - Bruno Berselli, Oct 03 2010
G.f.: x*(x^10 +2036*x^9 +152637*x^8 +2203488*x^7 +9738114*x^6 +15724248*x^5 +9738114*x^4 +2203488*x^3 +152637*x^2 +2036*x +1)/(1-x)^13. - Colin Barker, May 27 2012
a(n) = (-1)*Sum_{j=1..11} j*Stirling1(n+1,n+1-j)*Stirling2(n+11-j,n). - Mircea Merca, Jan 25 2014
a(n) = 1728*A006542(n+2)^2 + 216*A288876(n-2) + 96*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, May 25 2024

A132337 Sum of the integers from 1 to n, excluding the perfect sixth powers.

Original entry on oeis.org

0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484

Offset: 1

Cino Hilliard, Nov 07 2007

T. D. Noe, Table of n, a(n) for n = 1..10000

Different from A000096.

Cf. A132336, A178489.

A132337 := proc(n) r := floor(n^(1/6)) ; A000217(n)-A000540(r); end proc: seq(A132337(n),n=1..40) ; # R. J. Mathar

Accumulate[Table[If[IntegerQ[Surd[n,6]],0,n],{n,60}]] (* Harvey P. Dale, Jun 01 2022 *)

g6(n)=for(x=1,n,r=floor(x^(1/6));sum6=r^7/7+r^6/2+r^5/2-r^3/6+r/ 42;sn=x* (x+1)/2;print1(sn-sum6","))

A132337(n)=n*(n+1)/2-(1+n=floor(sqrtn(n+.5,6)))*(2*n+1)*((n^3+2*n^2-1)*n*3+1)*n/42 \\ M. F. Hasler, Oct 09 2010

Let r = floor(n^(1/6)). Then a(n) = n(n+1)/2 - (r^7/7 + r^6/2 + r^5/2 - r^3/6 + r/42) = A000217(n) - A000540(r).

a(n) = A000217(n) - A000540(A178489(n)). - M. F. Hasler, Oct 09 2010

Incorrect formula deleted by Jon E. Schoenfield, Jun 12 2010
Incorrect program replaced by R. J. Mathar, Oct 08 2010
Edited by the Assoc. Editors of the OEIS, Oct 12 2010. Thanks to Daniel Mondot for pointing out that the sequence needed editing.
Incorrect linear recurrence removed by Georg Fischer, Apr 11 2019

A181134 Sum of 13th powers: a(n) = Sum_{j=0..n} j^13.

Original entry on oeis.org

0, 1, 8193, 1602516, 68711380, 1289414505, 14350108521, 111239118928, 660994932816, 3202860761145, 13202860761145, 47725572905076, 154718778284148, 457593884876401, 1251308658130545, 3197503726489920

Offset: 0

Bruno Berselli, Oct 05 2010 - Oct 18 2010

This form of recurrence is a general property of the array in A103438 (sums of the first n-th powers).

Bruno Berselli, Table of n, a(n) for n = 0..10000.
Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian), 2008.
Index entries for linear recurrences with constant coefficients, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).

Cf. A010801.
Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), A023002 (m=10), A123095 (m=11), A123094 (m=12), A181134 (m=13).
Cf. A215083, A253712.

[(&+[j^13: j in [0..n]]): n in [0..30]]; // G. C. Greubel, Jul 21 2021

A181134 := proc(n) (bernoulli(14,n+1) - bernoulli(14))/14 ; end proc: seq(A181134(n), n=0..10); # R. J. Mathar, Oct 14 2010

Accumulate[Range[0,20]^13] (* Harvey P. Dale, Oct 30 2017 *)

A181134_list, m = [0], [6227020800, -37362124800, 97037740800, -142702560000, 130456085760, -76592355840, 28805736960, -6711344640, 901020120, -60780720, 1569750, -8190, 1, 0 , 0]
for _ in range(10**2):
    for i in range(14):
        m[i+1]+= m[i]
    A181134_list.append(m[-1]) # Chai Wah Wu, Nov 06 2014

[(bernoulli_polynomial(n+1, 14) - bernoulli(14))/14  for n in (0..30)] # G. C. Greubel, Jul 21 2021

For n>0, a(n) = n*A123094(n) - Sum_{i=0..n-1} A123094(i), where Sum_{i=0..n-1} A123094(i) = A253712(n-1) = (n-1)*n^2*(n+1)*(30*n^10 - 425*n^8 + 2578*n^6 - 8147*n^4 + 12874*n^2 - 7601)/5460.
a(n) = a(-n-1) = (n*(n + 1))^2*(30*n^10 + 150*n^9 + 125*n^8 - 400*n^7 - 326*n^6 + 1052*n^5 + 367*n^4 - 1786*n^3 + 202*n^2 + 1382*n - 691)/420.
G.f.: see comment of Vladeta Jovovic in A000538.
a(n) = -Sum_{j=1..13} j*Stirling1(n+1,n+1-j)*Stirling2(n+13-j,n). - Mircea Merca, Jan 25 2014

A081175 Numbers of the form Sum_{i=1..k} i^j, j >= 1, k >= 1.

Original entry on oeis.org

1, 3, 5, 6, 9, 10, 14, 15, 17, 21, 28, 30, 33, 36, 45, 55, 65, 66, 78, 91, 98, 100, 105, 120, 129, 136, 140, 153, 171, 190, 204, 210, 225, 231, 253, 257, 276, 285, 300, 325, 351, 354, 378, 385, 406, 435, 441, 465, 496, 506, 513, 528, 561, 595, 630, 650, 666, 703

Offset: 1

N. J. A. Sloane, Apr 18 2003

Union of sums of k-th powers, for k >= 1.

			30 is in the set because 30 = 1^2 + 2^2 + 3^2 + 4^2 (j=2, k=4).

Robert Israel, Table of n, a(n) for n = 1..10000
Michael Penn, an excruciatingly deep dive into the power sum., YouTube video, 2022.

Cf. A000217, A000330, A000537, A000538, A000539, A000540, A000541, A000542, A007487, A023002.

For primes in this sequence see A164307.

N:= 1000: # to get all terms <= N
A:=select(`<=`,{1, seq(seq(sum(i^k,i=1..m), m=2..floor((N*(k+1))^(1/(k+1)))),k = 1 ..ilog2(N-1))},N):
sort(convert(A,list)); # Robert Israel, Jan 26 2015

Take[ Union[ Flatten[ Table[ Sum[ i^j, {i, 1, n}], {j, 1, 9}, {n, 1, 40}]]], 60]

Corrected and extended by Robert G. Wilson v, May 08 2003

A181475 a(n) = 3n^4 + 6n^3 - 3*n + 1.

Original entry on oeis.org

1, 7, 91, 397, 1141, 2611, 5167, 9241, 15337, 24031, 35971, 51877, 72541, 98827, 131671, 172081, 221137, 279991, 349867, 432061, 527941, 638947, 766591, 912457, 1078201, 1265551, 1476307, 1712341, 1975597, 2268091, 2591911, 2949217, 3342241, 3773287, 4244731

Offset: 0

Bruno Berselli, Oct 25 2010 - Oct 29 2010

If gcd(n,7) = gcd(n+1,7) = gcd(2*n+1,7) = 1 then a(n) == 0 (mod 7) (E. Picutti, see References).

Ettore Picutti, Sul numero e la sua storia, Feltrinelli Economica, 1977, p. 208.

Bruno Berselli, Table of n, a(n) for n = 0..10000.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Subsequence of A003215.

Magma
```
[3*n^4+6*n^3-3*n+1: n in [0..31]];
```

Table[3 n^4 + 6 n^3 - 3 n + 1, {n, 0, 40}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,7,91,397,1141},40] (* Harvey P. Dale, Jul 12 2022 *)

G.f.: (1 + 2*x + 66*x^2 + 2*x^3 + x^4)/(1-x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 6*12.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6*A008594(n-1).
a(n) = 2*a(n-1) - a(n-2) + 6*A003154(n).
a(n) = a(n-1) + 6*A007588(n).
a(n) = 1 + 6*A062392(n).
a(n) = 7*A000540(n)/A000330(n) = A154105(A000096(n-1)) for n > 0.
Sum_{i=0..n} a(i) = (3*n^5 + 15*n^4 + 20*n^3 - 3*n + 5)/5.
a(n) = 7*(3*n^2 + 3*n - 1)*(Sum_{k=1..n} k^6)/(5*Sum_{k=1..n} k^4), n > 0. - Gary Detlefs, Oct 18 2011

Formula, program and crossref added by Bruno Berselli, Aug 22 2011

A079618 Triangle of coefficients in polynomials for partial sums of powers, scaled to produce integers: Sum_{i=1..m} i^(n-1) = Sum_{k=1..n} T(n,k)*m^k/A064538(n-1).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 0, 1, 2, 1, -1, 0, 10, 15, 6, 0, -1, 0, 5, 6, 2, 1, 0, -7, 0, 21, 21, 6, 0, 2, 0, -7, 0, 14, 12, 3, -3, 0, 20, 0, -42, 0, 60, 45, 10, 0, -3, 0, 10, 0, -14, 0, 15, 10, 2, 5, 0, -33, 0, 66, 0, -66, 0, 55, 33, 6, 0, 10, 0, -33, 0, 44, 0, -33, 0, 22, 12, 2, -691, 0, 4550, 0, -9009, 0, 8580, 0, -5005, 0, 2730, 1365, 210

Offset: 1

Henry Bottomley, Jan 29 2003

Rosinger connects this sequence to Weisstein's Faulhaber's Formula page. Rosinger also discusses, without reference to OEIS, (1.1) A000217 Triangular numbers: a(n) = C(n+1,2) = n*(n+1)/2 = 0+1+2+...+n; (1.2) A000330 Square pyramidal numbers: 0^2+1^2+2^2+...+n^2 = n*(n+1)*(2n+1)/6; (1.4) A033312 n! - 1 [with different offset and the formula 1*1! + 2*2! + 3*3! + ...]; (1.4) A007489 Sum_{k=1..n} k!. - Jonathan Vos Post, Feb 22 2007

			Triangle T(n, k) begins:
n\k 1   2   3   4   5    6   7   8   9 10 ...
1:  1
2:  1   1
3:  1   3   2
4:  0   1   2   1
5: -1   0  10  15   6
6:  0  -1   0   5   6    2
7:  1   0  -7   0  21   21   6
8:  0   2   0  -7   0   14  12   3
9: -3   0  20   0 -42    0  60  45  10
10: 0  -3   0  10   0  -14   0  15  10  2
... Reformatted. - _Wolfdieter Lang_, Feb 02 2015
For example row n=7: partial sums of 6th powers (A000540)
  1^6+2^6+...+m^6 = (m-7*m^3+21*m^5+21*m^6+6*m^7)/42.

Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, p. 106, 1996.

R. Mestrovic, A congruence modulo n^3 involving two consecutive sums of powers and its applications, arXiv:1211.4570 [math.NT], 2012. - From _N. J. A. Sloane_, Jan 03 2013
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
Elemer E. Rosinger, Synthesizing Sums, arXiv:math/0702605 [math.GM], 2007.
Eric Weisstein's World of Mathematics, Power Sum
Eric Weisstein's World of Mathematics, Faulhaber's Formula.

Cf. A064538, A000217, A000330, A007489, A033312.

T := proc(n, k) option remember; local A, B;
A := proc(n) option remember; denom((bernoulli(n+1,x)-bernoulli(n+1))/(n+1)) end:
B := proc(n) option remember; add(T(n,j),j=2..n) end;
if k>1 then T(n-1,k-1)*(n-1)*A(n-1)/(k*A(n-2)) elif n>1 then A(n-1)-B(n) else 1 fi end: seq(print(seq(T(n,k),k=1..n)),n=1..10); # Peter Luschny, Feb 02 2015
# Alternative:
A079618row := proc(n) bernoulli(n,x); (subs(x=x+1,%)-subs(x=1,%))/n;
seq(coeff(numer(%),x,k), k=1..n) end:
seq(A079618row(n), n=1..13); # Peter Luschny, Jul 14 2020

T[n_, k_] := T[n, k] = Module[{A, B}, A[m_] := A[m] = Denominator[ Together[ (BernoulliB[m+1, x] - BernoulliB[m+1])/(m+1)]]; B[m_] := B[m] = Sum[T[m, j], {j, 2, m}]; Which[k>1, T[n-1, k-1]*(n-1)*A[n-1]/(k*A[n-2]), n>1, A[n-1] - B[n], True, 1]]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 10}] // Flatten (* Jean-François Alcover, Sep 04 2015, after Peter Luschny *)

row(p) = {v = vector(p+1, k, (-1)^(k==p)*binomial(p+1, k)*bernfrac(p+1-k))/(p+1); lcmd = lcm(vector(#v, k, denominator(v[k]))); v*lcmd;}
tabl(nn) = for (n=0, nn, print(row(n))); \\ Michel Marcus, Feb 16 2016

T(n, k) = T(n-1, k-1) * (n-1) * A064538(n-1) / (k*A064538(n-2)) for k>1; T(n, 1) = A064538(n-1) - Sum_{k=2..n} T(n, k) for n>1; T(1, 1)=1.

Edited. Offset corrected from 0 to 1. Typo in formula corrected. - Wolfdieter Lang, Feb 02 2015

cp's OEIS Frontend

A123095 Sum of first n 11th powers.

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A132337 Sum of the integers from 1 to n, excluding the perfect sixth powers.

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A181134 Sum of 13th powers: a(n) = Sum_{j=0..n} j^13.

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A081175 Numbers of the form Sum_{i=1..k} i^j, j >= 1, k >= 1.

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A181475 a(n) = 3*n^4 + 6*n^3 - 3*n + 1.

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A079618 Triangle of coefficients in polynomials for partial sums of powers, scaled to produce integers: Sum_{i=1..m} i^(n-1) = Sum_{k=1..n} T(n,k)*m^k/A064538(n-1).

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A181475 a(n) = 3n^4 + 6n^3 - 3*n + 1.