A001014 -id:A001014 - cp's OEIS frontend

A254640 Third partial sums of sixth powers (A001014).

1, 67, 927, 6677, 32942, 126378, 404634, 1129854, 2833479, 6515509, 13947505, 28115451, 53846156, 98669156, 173975076, 296541132, 490504893, 789878583, 1241708083, 1909993393, 2880500634, 4266609710, 6216356510, 8920844010, 12624212835, 17635378761

Offset: 1

Luciano Ancora, Feb 04 2015

This is one of a sequence of arrays that are the convolutions of the zero-padded sequences binomial(2n-1+k,k) with the Eulerian polynomials E(n,x) of A008292, represented by E(n,x) (1-x)^(-2n), which generate increasing partial sums of powers of integers:
n= 2) (1 + 4*x + x^2)/(1-x)^4 is the o.g.f. of A000578, the convolution of (1,4,1) with A000292, giving the powers of m^3.
n= 3) (1 + 11*x + 11*x^2 + x^3)/(1-x)^6 is the o.g.f. of A000538, convolution of (1,11,11,1) with A000389, giving the partial sums of m^4.
n= 4) (1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1-x)^8, the o.g.f. of A101092, convolution of (1,26,66,26,1) with A000580, the second partial sums of m^5
n= 5) (1 + 57*x + 302*x^2 + 302*x^3 + 57*x^4 + x^5)/(1-x)^10, the o.g.f. of A254460, convolution of (1,57,302,302,57,1) with A000582, giving the third partial sums of m^6. - Tom Copeland, Dec 07 2015

Colin Barker, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A008292, A000578, A000292, A000538, A000389, A101092, A000580, A254460, A000582.

List([1..30], n-> Binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210); # G. C. Greubel, Aug 28 2019

[n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+ 5*n^4)/5040: n in [1..30]]; // Vincenzo Librandi, Feb 05 2015

seq(binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n(1+n)(2+n)(3+n)(3+2n)(2 -30n +35n^2 +30n^3 +5n^4)/5040, {n, 30}] (* or *) CoefficientList[Series[(x+1)(x^4 +56x^3 +246x^2 +56x +1)/(x - 1)^10, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 05 2015 *)

vector(30, n, n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+5*n^4)/5040) \\ Colin Barker, Feb 04 2015

def A254640(n): return n*(n*(n*(n*(n*(n*(n*(n*(10*n + 135) + 720) + 1890) + 2394) + 945) - 640) - 450) + 36)//5040 # Chai Wah Wu, Dec 07 2021

[binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210 for n in (1..30)] # G. C. Greubel, Aug 28 2019

a(n) = n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2 -30*n +35*n^2 +30*n^3 +5*n^4)/5040.

G.f.: x*(1+x)*(1 +56*x +246*x^2 +56*x^3 +x^4)/(1-x)^10. - Colin Barker, Feb 04 2015

A101093 Second partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 66, 860, 5750, 26265, 93436, 278256, 725220, 1703625, 3682030, 7431996, 14167946, 25730705, 44823000, 75305920, 122566056, 193963761, 299373690, 451829500, 668285310, 970507241, 1386109076, 1949746800, 2704487500

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Luciano Ancora, Recurrence relation for the second partial sums of m-th powers
Luciano Ancora, Second partial sums of the m-th powers
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 12-13.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000540.

List([2..30], n-> n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168); # G. C. Greubel, Aug 28 2019

[n*(1+n)^2*(2+n)*(-1+n*(2+n))*(-2+3*n*(2+n))/168: n in [1..40]]; // Vincenzo Librandi, Mar 24 2014

f:=n->(3*n^8-14*n^6+21*n^4-10*n^2)/168;
[seq(f(n),n=0..50)];  # N. J. A. Sloane, Mar 23 2014

CoefficientList[Series[(x+1)(x^4+56x^3+246x^2+56x+1)/(1-x)^9, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
Nest[Accumulate,Range[30]^6,2] (* or *) LinearRecurrence[{9,-36,84,-126, 126,-84,36,-9,1},{1,66,860,5750,26265,93436,278256,725220,1703625},30] (* Harvey P. Dale, Jun 05 2019 *)

vector(30, n, m=n+1; m^2*(3*m^6 -14*m^4 +21*m^2 -10)/168) \\ G. C. Greubel, Aug 28 2019

[n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168 for n in (2..30)] # G. C. Greubel, Aug 28 2019

a(n) = n*(1 + n)^2*(2 + n)*(-1 + n*(2 + n))*(-2 + 3*n*(2 + n))/168.
G.f.: x*(1+x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4)/(1-x)^9. - Colin Barker, Dec 18 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^6, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^6. - Luciano Ancora, Jan 08 2015

Edited by Ralf Stephan, Dec 16 2004

A254645 Fourth partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 68, 995, 7672, 40614, 166992, 571626, 1701480, 4534959, 11050468, 24997973, 53113424, 106959580, 205628736, 379603812, 676144944, 1166649837, 1956528420, 3198236503, 5108229896, 7988730530, 12255340240

Offset: 1

Luciano Ancora, Feb 05 2015

			First differences:   1, 63, 665, 3367, 11529,  31031, ...  (A022522)
--------------------------------------------------------------------------
The sixth powers:    1, 64, 729, 4096, 15625,  46656, ...  (A001014)
--------------------------------------------------------------------------
First partial sums:  1, 65, 794, 4890, 20515,  67171, ...  (A000540)
Second partial sums: 1, 66, 860, 5750, 26265,  93436, ...  (A101093)
Third partial sums:  1, 67, 927, 6677, 32942, 126378, ...  (A101099)
Fourth partial sums: 1, 68, 995, 7672, 40614, 166992, ...  (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000540, A001014, A022522, A101093, A101099.

Cf. A254644 (fourth partial sums of fifth powers), A254646 (fourth partial sums of seventh powers).

List([1..30], n-> Binomial(n+4,5)*(n+2)*((n^2+4*n-1)^2-2)/42); # G. C. Greubel, Aug 28 2019

[Binomial(n+4,5)*(n+2)*((n^2+4*n-1)^2-2)/42: n in [1..30]]; // G. C. Greubel, Aug 28 2019

seq(binomial(n+4,5)*(n+2)*((n^2+4*n-1)^2-2)/42, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n (1 + n) (2 + n)^2 (3 + n) (4 + n) (- 1 - 8 n + 14 n^2 + 8 n^3 + n^4)/5040, {n, 22}] (* or *)
Accumulate[Accumulate[Accumulate[Accumulate[Range[22]^6]]]] (* or *)
CoefficientList[Series[(- 1 - 57 x - 302 x^2 - 302 x^3 - 57 x^4 - x^5)/(- 1 + x)^11, {x, 0, 21}], x]
Nest[Accumulate,Range[30]^6,4] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,68,995,7672,40614,166992,571626,1701480,4534959,11050468,24997973},30] (* Harvey P. Dale, Dec 27 2015 *)

vector(30, n, binomial(n+4,5)*(n+2)*((n^2+4*n-1)^2-2)/42) \\ G. C. Greubel, Aug 28 2019

[binomial(n+4,5)*(n+2)*((n^2+4*n-1)^2-2)/42 for n in (1..30)] # G. C. Greubel, Aug 28 2019

G.f.: x*(1 + 57*x + 302*x^2 + 302*x^3 + 57*x^4 + x^5)/(1 - x)^11.
a(n) = n*(1 + n)*(2 + n)^2*(3 + n)*(4 + n)*(- 1 - 8*n + 14*n^2 + 8*n^3 + n^4)/5040.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + n^6.

A254683 Fifth partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 69, 1064, 8736, 49350, 216342, 787968, 2489448, 7024407, 18074875, 43072848, 96186272, 203145852, 408774588, 788378400, 1464523344, 2631173181, 4587701601, 7785938104, 12894168000, 20882898530, 33138238770

Offset: 1

Luciano Ancora, Feb 12 2015

			First differences:   1, 63,  665, 3367, 11529, ...  (A022522)
--------------------------------------------------------------------------
The sixth powers:    1, 64,  729, 4096, 15625, ...  (A001014)
--------------------------------------------------------------------------
First partial sums:  1, 65,  794, 4890, 20515, ...  (A000540)
Second partial sums: 1, 66,  860, 5750, 26265, ...  (A101093)
Third partial sums:  1, 67,  927, 6677, 32942, ...  (A254640)
Fourth partial sums: 1, 68,  995, 7672, 40614, ...  (A254645)
Fifth partial sums:  1, 69, 1064, 8736, 49350, ...  (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A000540, A001014, A022522, A101093, A254640, A254645, A254681, A254682, A254684.

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (5 + 2*n) (- 3 + 5*n + n^2) (4 + 15 n + 3 n^2)/332640, {n,22}] (* or *)
CoefficientList[Series[(1 + 57 x + 302 x^2 + 302 x^3 + 57 x^4 + x^5)/(- 1 + x)^12, {x,0,21}], x]

G.f.: (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)/(- 1 + x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(5 + 2*n)*(- 3 + 5*n + n^2)*(4 + 15*n + 3*n^2)/332640.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^6.

A254472 Sixth partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 70, 1134, 9870, 59220, 275562, 1063530, 3552978, 10577385, 28652260, 71725108, 167911380, 371057232, 779831820, 1568210220, 3032733564, 5663906745, 10251608346, 18037546450, 30931714450, 51814612980, 84952851750, 136562787270, 215565263550, 334584493425

Offset: 1

Luciano Ancora, Feb 15 2015

			First differences:   1, 63,  665, 3367, 11529, ... (A022522)
--------------------------------------------------------------------------
The sixth powers:    1, 64,  729, 4096, 15625, ... (A001014)
--------------------------------------------------------------------------
First partial sums:  1, 65,  794, 4890, 20515, ... (A000540)
Second partial sums: 1, 66,  860, 5750, 26265, ... (A101093)
Third partial sums:  1, 67,  927, 6677, 32942, ... (A254640)
Fourth partial sums: 1, 68,  995, 7672, 40614, ... (A254645)
Fifth partial sums:  1, 69, 1064, 8736, 49350, ... (A254683)
Sixth partial sums:  1, 70, 1134, 9870, 59220, ... (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers.
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A000540, A001014, A022522, A101093, A254469, A254470, A254471, A254640, A254645, A254683.

[n*(1+n)*(2+n)*(3+n)^2*(4+n)*(5+n)*(6+n)*(-3+5*n+n^2)* (3+7*n+n^2)/665280: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n) (2 + n) (3 + n)^2 (4 + n) (5 + n) (6 + n) (- 3 + 5 n + n^2) (3 + 7 n + n^2)/665280, {n, 22}] (* or *) CoefficientList[Series[(- 1 - 57 x - 302 x^2 - 302 x^3 - 57 x^4 - x^5)/(- 1 + x)^13, {x, 0, 28}], x]
Nest[Accumulate,Range[30]^6,6] (* Harvey P. Dale, Oct 02 2015 *)

vector(50,n,n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(-3 + 5*n + n^2)*(3 + 7*n + n^2)/665280) \\ Derek Orr, Feb 19 2015

G.f.: (-x - 57*x^2 - 302*x^3 - 302*x^4 - 57*x^5 - x^6)/(- 1 + x)^13.
a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(-3 + 5*n + n^2)*(3 + 7*n + n^2)/665280.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + n^6.
Sum_{n>=1} 1/a(n) = 25622179/76545 - 3080*Pi^2/81 + 149600*Pi*tan(sqrt(37)*Pi/2)/(243*sqrt(37)). - Amiram Eldar, Jan 27 2022

A254872 Seventh partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 71, 1205, 11075, 70295, 345857, 1409387, 4962365, 15539750, 44192010, 115917118, 283828498, 654885730, 1434717550, 3002927770, 6035661334, 11699568079, 21951176425, 39988722875, 70920437325, 122735050305

Offset: 1

Luciano Ancora, Feb 17 2015

			First differences:    1, 63,  665,  3367, 11529, ... (A022522)
--------------------------------------------------------------------
The sixth powers:     1, 64,  729,  4096, 15625, ... (A001014)
--------------------------------------------------------------------
First partial sums:   1, 65,  794,  4890, 20515, ... (A000540)
Second partial sums:  1, 66,  860,  5750, 26265, ... (A101093)
Third partial sums:   1, 67,  927,  6677, 32942, ... (A254640)
Fourth partial sums:  1, 68,  995,  7672, 40614, ... (A254645)
Fifth partial sums:   1, 69, 1064,  8736, 49350, ... (A254683)
Sixth partial sums:   1, 70, 1134,  9870, 59220, ... (A254472)
Seventh partial sums: 1, 71, 1205, 11075, 70295, ... (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Cf. A000540, A001014, A022522, A101093, A254472, A254640, A254645, A254683, A254869, A254870, A254871.

Table[(n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (7 + 2 n) (- 49 + 147 n^2 + 42 n^3 + 3 n^4))/51891840, {n, 21}] (* or *)
CoefficientList[Series[(1 + 57 x + 302 x^2 + 302 x^3 + 57 x^4 + x^5)/(- 1 + x)^14, {x, 0, 20}], x]

G.f.: (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)/(- 1 + x)^14.
a(n) = (n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 2*n)*(- 49 + 147*n^2 + 42*n^3 + 3*n^4))/51891840.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + n^6.

A279639 Exponential transform of the sixth powers A001014.

Original entry on oeis.org

1, 1, 65, 922, 19685, 572036, 16379797, 542459296, 20028938953, 787480005520, 33447797179721, 1522102664036384, 73362723948758125, 3738119667151161280, 200625910519541044189, 11290451562860730241216, 664399657108812332697233, 40781390340823661046136064

Offset: 0

Alois P. Heinz, Dec 16 2016

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..405

Column k=6 of A279636.

Cf. A001014.

a:= proc(n) option remember; `if`(n=0, 1,
      add(binomial(n-1, j-1)*j^6*a(n-j), j=1..n))
    end:
seq(a(n), n=0..25);

E.g.f.: exp(exp(x)*(x^6+15*x^5+65*x^4+90*x^3+31*x^2+x)).

A268335 Exponentially odd numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 88, 89, 91, 93, 94, 95, 96, 97

Offset: 1

Vladimir Shevelev, Feb 01 2016

nonn

The sequence is formed by 1 and the numbers whose prime power factorization contains only odd exponents.
The density of the sequence is the constant given by A065463.
Except for the first term the same as A002035. - R. J. Mathar, Feb 07 2016
Also numbers k all of whose divisors are bi-unitary divisors (i.e., A286324(k) = A000005(k)). - Amiram Eldar, Dec 19 2018
The term "exponentially odd integers" was apparently coined by Cohen (1960). These numbers were also called "unitarily 2-free", or "2-skew", by Cohen (1961). - Amiram Eldar, Jan 22 2024

Peter J. C. Moses, Table of n, a(n) for n = 1..2000
Eckford Cohen, Arithmetical functions associated with the unitary divisors of an integer, Mathematische Zeitschrift, Vol. 74 (1960), pp. 66-80.
Eckford Cohen, Some sets of integers related to the k-free integers, Acta Sci. Math. (Szeged), Vol. 22, No. 3-4 (1961), pp. 223-233.
Vladimir Shevelev, Exponentially S-numbers, arXiv:1510.05914 [math.NT], 2015.
Vladimir Shevelev, Set of all densities of exponentially S-numbers, arXiv preprint arXiv:1511.03860 [math.NT], 2015.
Vladimir Shevelev, S-exponential numbers, Acta Arithmetica, Vol. 175(2016), 385-395.
D. Suryanarayana and R. Sita Rama Chandra Rao, Distribution of unitarily k-free integers, Journal of the Australian Mathematical Society, Vol. 20 , No. 2 (1975), pp. 129-141.

Cf. A002035, A209061, A138302, A197680, A000578, A000584, A001014, A001017, A008456, A010803, A010805, A010806, A010808, A010811, A010812, A001221, A124010.

Select[Range@ 100, AllTrue[Last /@ FactorInteger@ #, OddQ] &] (* Version 10, or *)
Select[Range@ 100, Times @@ Boole[OddQ /@ Last /@ FactorInteger@ #] == 1 &] (* Michael De Vlieger, Feb 02 2016 *)

isok(n)=my(f = factor(n)); for (k=1, #f~, if (!(f[k,2] % 2), return (0))); 1; \\ Michel Marcus, Feb 02 2016

from itertools import count, islice
from sympy import factorint
def A268335_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:all(e&1 for e in factorint(n).values()),count(max(startvalue,1)))
A268335_list = list(islice(A268335_gen(),20)) # Chai Wah Wu, Jun 22 2023

Sum_{a(n)<=x} 1 = C*x + O(sqrt(x)*log x*e^(c*sqrt(log x)/(log(log x))), where c = 4*sqrt(2.4/log 2) = 7.44308... and C = Product_{prime p} (1 - 1/p*(p + 1)) = 0.7044422009991... (A065463).

Sum_{n>=1} 1/a(n)^s = zeta(2*s) * Product_{p prime} (1 + 1/p^s - 1/p^(2*s)), s>1. - Amiram Eldar, Sep 26 2023

A013954 a(n) = sigma_6(n), the sum of the 6th powers of the divisors of n.

Original entry on oeis.org

1, 65, 730, 4161, 15626, 47450, 117650, 266305, 532171, 1015690, 1771562, 3037530, 4826810, 7647250, 11406980, 17043521, 24137570, 34591115, 47045882, 65019786, 85884500, 115151530, 148035890, 194402650, 244156251, 313742650, 387952660, 489541650, 594823322, 741453700

Offset: 1

N. J. A. Sloane

If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
Inverse Mobius transform of A001014. - R. J. Mathar, Oct 13 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for sequences related to sigma(n).

Cf. A000203, A001014, A001157-A001160, A013665, A013955-A013972, A017665-A017712.

[DivisorSigma(6,n): n in [1..30]]; // Bruno Berselli, Apr 10 2013

A013954 := proc(n)
        numtheory[sigma][6](n) ;
end proc: # R. J. Mathar, Oct 13 2011

lst={};Do[AppendTo[lst,DivisorSigma[6,n]],{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)
DivisorSigma[6,Range[30]] (* Harvey P. Dale, May 11 2025 *)

a(n)=sigma(n,6) \\ Charles R Greathouse IV, Apr 28 2011

[sigma(n,6)for n in range(1,24)] # Zerinvary Lajos, Jun 04 2009

G.f.: Sum_{k>=1} k^6*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
L.g.f.: -log(Product_{k>=1} (1 - x^k)^(k^5)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 06 2017
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(6*e+6)-1)/(p^6-1).
Dirichlet g.f.: zeta(s)*zeta(s-6).
Sum_{k=1..n} a(k) = zeta(7) * n^7 / 7 + O(n^8). (End)

A001110 Square triangular numbers: numbers that are both triangular and square.

Original entry on oeis.org

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100

Offset: 0

N. J. A. Sloane

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004
For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007
The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012
Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012
For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013
For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014
The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014
The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

			a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 204.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-259.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.

Kade Robertson, Table of n, a(n) for n = 0..600 [This replaces an earlier b-file computed by T. D. Noe]
Nikola Adžaga, Andrej Dujella, Dijana Kreso, and Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares, The Mathematical Gazette, 2002, pp. 423-431, esp pp. 424-426.
K. S. Brown, Square Triangular Numbers.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
Yong-Gao Chen and Jin-Hui Fang, Triangular numbers in geometric progression, INTEGERS 7 (2007), #A19.
F. Javier de Vega, On the parabolic partitions of a number, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
Colin Dickson et al., ratio of integers = sqrt(2), thread in newsgroup alt.math.recreational, March 7, 2004.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
Bill Gosper, The Triangular Squares, 2014.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020. Mentions this sequence.
Gillian Hatch, Pythagorean Triples and Triangular Square Numbers, Mathematical Gazette 79 (1995), 51-55.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Roger B. Nelsen, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
A. Nowicki, The numbers a^2+b^2-dc^2, J. Int. Seq. 18 (2015) # 15.2.3.
Matthew Parker, The first 5000 square triangular numbers (7-Zip compressed file).
J. L. Pietenpol, A. V. Sylwester, E. Just, and R. M. Warten, Problem E1473, Amer. Math. Monthly, Vol. 69, No. 2 (Feb. 1962), pp. 168-169. (From the editorial note on p. 169 of this source, we learn that the question about the existence of perfect squares in the sequence of triangular numbers cropped up in the Euler-Goldbach Briefwechsel of 1730; the translation into English of the relevant letters can be found at Correspondence of Leonhard Euler with Christian Goldbach (part II), pp. 614-615.) - _José Hernández_, May 24 2022
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
D. A. Q., Triangular square numbers: a postscript, Math. Gaz., 56 (1972), 311-314.
K. Ramsey, Re_Generalized_Proof_re_Square_Triangular_Numbers. [Broken link]
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Fabio Roman, On certain ratios regarding integer numbers which are both triangulars and squares, arXiv:1703.06701 [math.NT], 2017.
Jon E. Schoenfield, Prime factorization of a(n) for n = 1..300.
Jaap Spies, A Bit of Math, The Art of Problem Solving, Jaap Spies Publishers (2019).
Ralf Stephan, Boring proof of a nonlinearity.
UWC, Problem A, Nieuw Archief voor Wiskunde, Dec 2004; Jaap Spies, Solution.
Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Square triangular number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001108, A001109.
Other S_r type sequences are S_4=A000290, S_5=A004146, S_7=A054493, S_8=A001108, S_9=A049684, S_20=A049683, S_36=this sequence, S_49=A049682, S_144=A004191^2.
Cf. A001014; intersection of A000217 and A000290; A010052(a(n))*A010054(a(n)) = 1.
Cf. A005214, A054686, A232847 and also A233267 (reveals an interesting divisibility pattern for this sequence).
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
See A229131, A182334, A299921 for near-misses.

a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011

[n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015

a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq(A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); # Simon Plouffe in his 1992 dissertation

f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]],AppendTo[lst,f[n]]],{n,0,10!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
Transpose[NestList[Flatten[{Rest[#],34Last[#]-First[#]+2}]&, {0,1},20]][[1]]  (* Harvey P. Dale, Mar 25 2011 *)
LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *)
LinearRecurrence[{6,-1},{0,1},20]^2 (* Harvey P. Dale, Oct 22 2012 *)
(* Square = Triangular = Triangular = A001110 *)
ChebyshevU[#-1,3]^2==Binomial[ChebyshevT[#/2,3]^2,2]==Binomial[(1+ChebyshevT[#,3])/2,2]=={1,36,1225,41616,1413721}[[#]]&@Range[5]
True (* Bill Gosper, Jul 20 2015 *)
L=0;r={};Do[AppendTo[r,L];L=1+17*L+6*Sqrt[L+8*L^2],{i,1,19}];r (* Kebbaj Mohamed Reda, Aug 02 2023 *)

a=vector(100);a[1]=1;a[2]=36;for(n=3,#a,a[n]=34*a[n-1]-a[n-2]+2);a \\ Charles R Greathouse IV, Jul 25 2011

;; With memoizing definec-macro from Antti Karttunen's IntSeq-library.
(definec (A001110 n) (if (< n 2) n (+ 2 (- (* 34 (A001110 (- n 1))) (A001110 (- n 2))))))
;; Antti Karttunen, Dec 06 2013

;; For testing whether n is in this sequence:
(define (inA001110? n) (and (zero? (A068527 n)) (inA001109? (floor->exact (sqrt n)))))
(define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
;; Antti Karttunen, Dec 06 2013

a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001
a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006
a(n) = A001109(n)^2 = A001108(n)*(A001108(n)+1)/2 = (A000129(n)*A001333(n))^2 = (A000129(n)*(A000129(n) + A000129(n-1)))^2. - Henry Bottomley, Apr 19 2000
a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004
From Antonio Alberto Olivares, Nov 07 2003: (Start)
a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
a(n+1) = (17*A029547(n) - A091761(n) - 1)/16. - R. J. Mathar, Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129(2*n)^2/4 = binomial(A001108,2). - Bill Gosper, Jul 28 2008
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008
a(n) = (1/8)*(sinh(2*n*arcsinh(1)))^2. - Artur Jasinski, Feb 10 2010
a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011
a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011
a(2*n+1) = A002315(n)^2*(A002315(n)^2 + 1)/2. - Ivan N. Ianakiev, Oct 10 2012
a(2*n+1) = ((sqrt(t^2 + (t+1)^2))*(2*t+1))^2, where t = (A002315(n) - 1)/2. - Ivan N. Ianakiev, Nov 01 2012
a(2*n) = A001333(2*n)^2 * (A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * (A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013
a(n) = (A001542(n)/2)^2.
From Peter Bala, Apr 03 2014: (Start)
a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = A096979(2*n-1) for n > 0. - Ivan N. Ianakiev, Jun 21 2014
a(n) = (6*sqrt(a(n-1)) - sqrt(a(n-2)))^2. - Arkadiusz Wesolowski, Apr 06 2015
From Daniel Poveda Parrilla, Jul 16 2016 and Sep 21 2016: (Start)
a(n) = A000290(A002965(2*n)*A002965(2*n + 1)) (after Hugh Darwen).
a(n) = A000217(2*(A000129(n))^2 - (A000129(n) mod 2)).
a(n) = A000129(n)^4 + Sum_{k=0..(A000129(n)^2 - (A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers).
a(n) = A002965(2*n)^4 + Sum_{k=A002965(2*n)^2..A002965(2*n)*A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End)
E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016

cp's OEIS Frontend

A254640 Third partial sums of sixth powers (A001014).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

A101093 Second partial sums of sixth powers (A001014).

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A254645 Fourth partial sums of sixth powers (A001014).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A254683 Fifth partial sums of sixth powers (A001014).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A254472 Sixth partial sums of sixth powers (A001014).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A254872 Seventh partial sums of sixth powers (A001014).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A279639 Exponential transform of the sixth powers A001014.