A000583 -id:A000583 - cp's OEIS frontend

A046042 Number of partitions of n into fourth powers.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 9, 9, 9, 9, 9, 9

Offset: 1

Eric W. Weisstein

nonn

In general, the number of partitions of n into perfect s-th powers (s>=1) is asymptotic to (2*Pi)^(-(s+1)/2) * sqrt(s/(s+1)) * k * n^(1/(s+1)-3/2) * exp((s+1)*k*n^(1/(s+1))), where k = (Gamma(1 + 1/s) * Zeta(1 + 1/s) / s)^(s/(s+1)) [Hardy & Ramanujan, 1917]. - Vaclav Kotesovec, Dec 29 2016

			a(33) = 3 because we have [16,16,1], [16,1,1,...,1] (17 1's) and [1,1,...,1] (33 1's).

H. P. Robinson, Letter to N. J. A. Sloane, Jan 04 1974.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
G. H. Hardy and S. Ramanujan, Asymptotic formulae in combinatory analysis, Proceedings of the London Mathematical Society, 2, XVI, 1917, p. 373.
Herman P. Robinson, Letter to N. J. A. Sloane, Jan 1974.
Eric Weisstein's World of Mathematics, Partition

Cf. A000583, A002377, A003105.
Cf. A001156, A003108, A046042.
Cf. A037444, A259792, A259793.

a046042 = p $ tail a000583_list where
   p _          0 = 1
   p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
-- Reinhard Zumkeller, May 18 2015   ~

g:=-1+1/product(1-x^(j^4),j=1..10): gser:=series(g,x=0,105): seq(coeff(gser,x,n),n=1..102); # Emeric Deutsch, Apr 06 2006

g = -1 + 1/Product[1 - x^(j^4), {j, 1, 10}]; gser =
Series[g, {x, 0, 105}]; Table[Coefficient[gser, x, n], {n, 1, 102}] (* Jean-François Alcover, Oct 29 2012, after Emeric Deutsch *)

G.f.: -1+1/product(1-x^(j^4),j=1..infinity). - Emeric Deutsch, Apr 06 2006
a(n) ~ exp(5 * (Gamma(1/4)*Zeta(5/4))^(4/5) * n^(1/5) / 2^(16/5)) * (Gamma(1/4)*Zeta(5/4))^(4/5) / (2^(47/10) * sqrt(5) * Pi^(5/2) * n^(13/10)) [Hardy & Ramanujan, 1917]. - Vaclav Kotesovec, Dec 29 2016
G.f.: Sum_{i>=1} x^(i^4) / Product_{j=1..i} (1 - x^(j^4)). - Ilya Gutkovskiy, May 07 2017

A262954 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y(y+1)/2 has the form z^2 or 8z^2.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 3, 2, 1, 4, 5, 3, 1, 2, 4, 3, 3, 3, 5, 4, 2, 4, 5, 2, 3, 6, 4, 3, 4, 5, 5, 4, 3, 2, 5, 4, 7, 7, 3, 4, 3, 4, 2, 4, 6, 6, 6, 2, 2, 2, 4, 5, 9, 5, 4, 5, 2, 3, 2, 5, 5, 5, 2, 4, 5, 3, 4, 5, 4, 5, 7, 3, 3, 3, 6, 3, 4, 4, 5, 6, 3, 7, 7, 3, 4, 8, 7, 7, 1, 3, 9, 8, 6

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 13, 89, 449, 524, 1049, 2164, 14969, 51334.
(ii) For any positive integer n, there are integers x and y > 0 such that n - x^4 - T(y) has the form T(z) or 4*T(z), where T(k) refers to the triangular number k*(k+1)/2.
(iii) For every n = 1,2,3,... there are integers x and y > 0 such that n - x^4 - T(y) has the form T(z) or 2*z^2.
(iv) For {c,d} = {1,2} and n > 0, there are integers x and y > 0 such that n - 2*x^4 - T(y) has the form c*T(z) or d*z^2.
(v) For each n = 1,2,3,... there are integers x and y > 0 such that n - 4*x^4 - T(y) has the form 2*T(z) or z^2.
See also A262941, A262944, A262945, A262954, A262955 and A262956 for similar conjectures.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + 0^2.
a(9) = 1 since 9 = 0^4 + 1*2/2 + 8*1^2.
a(13) = 1 since 13 = 1^4 + 2*3/2 + 3^2.
a(89) = 1 since 89 = 2^4 + 1*2/2 + 8*3^2.
a(449) = 1 since 449 = 0^4 + 22*23/2 + 14^2.
a(524) = 1 since 524 = 3^4 + 29*30/2 + 8*1^2.
a(1049) = 1 since 1049 = 5^4 + 16*17/2 + 8*6^2.
a(2164) = 1 since 2164 = 1^4 + 34*35/2 + 8*14^2.
a(14969) = 1 since 14969 = 8^4 + 145*146/2 + 8*6^2.
a(51334) = 1 since 51334 = 5^4 + 313*314/2 + 8*14^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000583, A262941, A262944, A262945, A262954, A262955, A262956.

SQ[n_]:=IntegerQ[Sqrt[n]]||IntegerQ[Sqrt[n/8]]
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A209593 T(n,k)=Number of n X n 0..k arrays with every element equal to a diagonal or antidiagonal reflection.

Original entry on oeis.org

2, 3, 16, 4, 81, 192, 5, 256, 3645, 9216, 6, 625, 28672, 1476225, 663552, 7, 1296, 140625, 51380224, 996451875, 191102976, 8, 2401, 513216, 791015625, 161128382464, 6053445140625, 82556485632, 9, 4096, 1529437, 7316407296

Offset: 1

R. H. Hardin Mar 10 2012

Table starts
......2.........3............4.............5...............6................7
.....16........81..........256...........625............1296.............2401
....192......3645........28672........140625..........513216..........1529437
...9216...1476225.....51380224.....791015625......7316407296......47738317081
.663552.996451875.161128382464.8009033203125.191221621088256.2767247026234327

			Some solutions for n=3 k=3
..2..0..1....1..0..1....0..0..3....0..2..1....1..2..3....0..1..2....2..2..0
..3..1..0....0..1..0....2..1..0....2..3..2....2..3..0....1..3..2....1..0..2
..0..3..3....1..0..1....1..2..3....3..2..3....2..0..2....2..2..2....1..1..3

R. H. Hardin, Table of n, a(n) for n = 1..210

Row 2 is A000583(n+1)

T(n,k) = (k+1) ^ (2n-(n modulo 2)) * ((k+1)*(2k+1)) ^ ((n*n-2n+(n modulo 2))/4)

A262944 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a pentagonal number.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 4, 3, 1, 3, 5, 3, 2, 2, 5, 5, 3, 3, 5, 5, 3, 6, 6, 3, 3, 8, 6, 5, 5, 3, 7, 5, 5, 3, 4, 4, 8, 9, 3, 5, 7, 6, 3, 5, 5, 7, 5, 3, 4, 5, 6, 6, 9, 4, 5, 7, 7, 5, 4, 4, 7, 6, 1, 5, 5, 7, 7, 7, 1, 6, 10, 8, 6, 3, 4, 3, 6, 4, 6, 9, 5, 7, 9, 3, 5, 8, 9, 8, 3, 3, 11, 10, 6, 6, 8, 12, 5, 6, 4, 7

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 63, 69, 489, 714, 1089.
(ii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is twice a square or twice a pentagonal number.
(iii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - 2*x^4 - y*(y+1)/2 is a square or a pentagonal number.
See also A262941 and A262945 for similar conjectures.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(9) = 1 since 9 = 1^4 + 2*3/2 + p_5(2).
a(63) = 1 since 63 = 0^4 + 7*8/2 + p_5(5).
a(69) = 1 since 69 = 2^4 + 7*8/2 + 5^2.
a(489) = 1 since 489 = 3^4 + 12*13/2 + p_5(15).
a(714) = 1 since 714 = 4^4 + 18*19/2 + p_5(14).
a(1089) = 1 since 1089 = 4^4 + 38*39/2 + p_5(8).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000217, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945.

SQ[n_]:=IntegerQ[Sqrt[n]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1,6]==0)
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A262945 Number of ordered pairs (x,y) with x >= 0 and y >= 0 such that n - x^4 - 2*y^2 is a triangular number or a pentagonal number.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 4, 5, 2, 5, 4, 5, 7, 3, 1, 1, 4, 4, 6, 4, 1, 4, 4, 3, 5, 6, 5, 6, 4, 1, 1, 2, 5, 4, 5, 3, 3, 2, 1, 5, 4, 7, 9, 5, 4, 2, 2, 2, 5, 3, 2, 5, 2, 1, 3, 4, 3, 8, 4, 4, 5, 6, 3, 3, 3, 2, 7, 6, 1, 3, 3, 4, 7, 4, 6, 6, 7, 5, 2, 3, 3

Offset: 0

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: a(n) > 0 for every n = 0,1,2,..., and a(n) = 1 only for the following 55 values of n: 0, 26, 27, 32, 41, 42, 50, 65, 80, 97, 112, 122, 130, 160, 196, 227, 239, 272, 322, 371, 612, 647, 736, 967, 995, 1007, 1106, 1127, 1205, 1237, 1240, 1262, 1637, 1657, 1757, 2912, 2987, 3062, 3107, 3524, 3647, 3902, 5387, 5587, 5657, 6047, 6107, 11462, 13427, 14717, 15002, 17132, 20462, 30082, 35750.

See also A262941, A262944, A262954 and A262955 for similar conjectures.

			a(26) = 1 since 26 = 2^4 + 2*0^2 + 4*5/2.
a(32) = 1 since 32 = 0^4 + 2*4^2 + 0*1/2.
a(41) = 1 since 41 = 1^4 + 2*3^2 + p_5(4), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(196) = 1 since 196 = 1^4 + 2*5^2 + p_5(10).
a(3524) = 1 since 3524 = 0^4 + 2*22^2 + 71*72/2.
a(3647) = 1 since 3647 = 0^4 + 2*34^2 + p_5(30).
a(6047) = 1 since 6047 = 5^4 + 2*39^2 + p_5(40).
a(6107) = 1 since 6107 = 0^4 + 2*1^2 + 110*111/2.
a(11462) = 1 since 11462 = 9^4 + 2*5^2 + 98*99/2.
a(13427) = 1 since 13427 = 7^4 + 2*0^2 + 148*149/2.
a(14717) = 1 since 14717 = 8^4 + 2*72^2 + 22*23/2.
a(15002) = 1 since 15002 = 0^4 + 2*86^2 + 20*21/2.
a(17132) = 1 since 17132 = 3^4 + 2*30^2 + p_5(101).
a(20462) = 1 since 20462 = 0^4 + 2*26^2 + 195*196/2.
a(30082) = 1 since 30082 = 11^4 + 2*63^2 + 122*123/2.
a(35750) = 1 since 35750 = 0^4 + 2*44^2 + 252*253/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000217, A000290, A000326, A000583, A262941, A262944, A262954, A262955.

SQ[n_]:=IntegerQ[Sqrt[8n+1]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1, 6]==0)
Do[r=0;Do[If[SQ[n-x^4-2y^2],r=r+1],{x, 0, n^(1/4)},{y,0,Sqrt[(n-x^4)/2]}];Print[n, " ", r];Continue,{n,0,100}]

A262956 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a square minus 1.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 2, 2, 5, 5, 3, 2, 3, 4, 4, 4, 5, 7, 5, 3, 6, 5, 3, 7, 8, 5, 4, 5, 7, 8, 6, 2, 4, 5, 5, 10, 7, 5, 7, 6, 4, 3, 5, 8, 10, 6, 2, 3, 5, 6, 10, 9, 5, 7, 6, 4, 4, 5, 6, 8, 5, 3, 8, 7, 5, 7, 5, 6, 11, 9, 5, 3, 5, 5, 4, 4, 3, 8, 9, 7, 10, 7, 5, 11, 10, 8, 5, 1

Offset: 1

Zhi-Wei Sun, Oct 05 2015

nonn

Conjecture: a(n) > 0 for all n > 0. In other words, for any positive integer n, either n or n + 1 can be written as the sum of a fourth power, a square and a positive triangular number.

We also guess that a(n) = 1 only for n = 1, 89, 244, 464, 5243, 14343.

			a(1) = 1 since 1 = 0^4 + 1*2/2 + 0^2.
a(89) = 1 since 89 = 2^4 + 4*5/2 + 8^2 - 1.
a(244) = 1 since 244 = 2^4 + 2*3/2 + 15^2.
a(464) = 1 since 464 = 2^4 + 22*23/2 + 14^2 - 1.
a(5243) = 1 since 5243 = 0^4 + 50*51/2 + 63^2 - 1.
a(14343) = 1 since 14343 = 2^4 + 163*164/2 + 31^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000583, A262941, A262944, A262945, A262954, A262955, A262959.

SQ[n_]:=IntegerQ[Sqrt[n]]||IntegerQ[Sqrt[n+1]]
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A266152 Least positive integer y such that n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists.

Original entry on oeis.org

8, 1, 2, 17, 1, 3, 139, 19, 37, 1, 3, 9, 2, 7, 3, 1411, 1, 2, 2, 1, 5, 4, 387, 3, 1, 1, 4, 7, 9, 2, 35, 1, 33, 2, 6, 5, 1, 4, 3, 11, 1, 6, 2, 429, 2, 5, 11, 179, 73, 1, 15, 1, 4, 3, 11, 3, 5, 2, 3, 15, 5, 6, 7, 3, 1, 6, 4, 6337, 8, 16, 3

Offset: 0

Zhi-Wei Sun, Dec 22 2015

nonn

Conjecture: Any integer m can be written as x^4 - y^3 + z^2, where x, y and z are positive integers.
This is slightly stronger than the conjecture in A266003.
See also A266153 for a related sequence, and A266212 for a stronger conjecture.
If n is a positive square, then a(n) = 1. - Altug Alkan, Dec 23 2015

			a(0) = 8 since 0 = 4^4 - 8^3 + 16^2.
a(6) = 139 since 6 = 36^4 - 139^3 + 1003^2.
a(15) = 1411 since 15 = 119^4 - 1411^3 + 51075^2.
a(11019) = 71383 since 11019 = 4325^4 - 71383^3 + 3719409^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Checking the conjecture for integers m with |m| <= 10^5
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000290, A000578, A000583, A262827, A266003, A266004, A266153, A266212.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
Do[y=1;Label[bb];Do[If[SQ[n+y^3-x^4],Print[n," ",y];Goto[aa]],{x,1,(n+y^3)^(1/4)}];y=y+1;Goto[bb];Label[aa];Continue,{n,0,70}]

A010801 13th powers: a(n) = n^13.

Original entry on oeis.org

0, 1, 8192, 1594323, 67108864, 1220703125, 13060694016, 96889010407, 549755813888, 2541865828329, 10000000000000, 34522712143931, 106993205379072, 302875106592253, 793714773254144, 1946195068359375, 4503599627370496, 9904578032905937, 20822964865671168

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Cf. A000290 (squares), A000578 (cubes), A000583 (4th powers), A000584 (5th powers), A008455 (11th powers), A013671 (zeta(11)).

[n^13: n in [0..15]]; // Vincenzo Librandi, Jun 19 2011

A010801 := n -> n^13; \\ M. F. Hasler, Jul 03 2025

Range[0,30]^13 (* Vladimir Joseph Stephan Orlovsky, Mar 14 2011 *)

A010801(n)=n^13 \\ Charles R Greathouse IV, Oct 07 2015

A010801 = lambda n: n**13 # M. F. Hasler, Jul 03 2025

a(n) mod 10 = n mod 10. - Reinhard Zumkeller, Dec 06 2004
Totally multiplicative with a(p) = p^13 for primes p. Multiplicative with a(p^e) = p^(13*e). - Jaroslav Krizek, Nov 01 2009
G.f.: x*(x^12 + 8178*x^11 + 1479726*x^10 + 45533450*x^9 + 423281535*x^8 + 1505621508*x^7 + 2275172004*x^6 + 1505621508*x^5 + 423281535*x^4 + 45533450*x^3 + 1479726*x^2 + 8178*x + 1) / (x - 1)^14. - Colin Barker, Sep 25 2014
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(13) (A013671).
Sum_{n>=1} (-1)^(n+1)/a(n) = 4095*zeta(13)/4096. (End)

A058895 a(n) = n^4 - n.

Original entry on oeis.org

0, 0, 14, 78, 252, 620, 1290, 2394, 4088, 6552, 9990, 14630, 20724, 28548, 38402, 50610, 65520, 83504, 104958, 130302, 159980, 194460, 234234, 279818, 331752, 390600, 456950, 531414, 614628, 707252, 809970, 923490, 1048544, 1185888, 1336302, 1500590, 1679580

Offset: 0

Henry Bottomley, Jan 08 2001

a(n) is the number of ways to assign 4 different students to n different dorm rooms, each of which can hold at most 3 students. In other words, a(n) is the number of functions f:[4]->[n] with the size of the pre-image set of each element of the codomain at most 3. - Dennis P. Walsh, Mar 21 2013

a(n) are the values of m that yield integer solutions to this family of equations: x = sqrt(m + sqrt(x)), which may also be viewed as an infinitely recursive radical. The real solutions for x at each m = a(n) is n^2, except at n = 1 (m = 0) where x = 0 or 1 is a solution. - Richard R. Forberg, Oct 15 2014

Harry J. Smith, Table of n, a(n) for n = 0..2000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A002061, A002378, A024001, A027444, A027482, A068601, A033562, A185065, A339605.

[n^4-n: n in [0..40]]; // Vincenzo Librandi, Feb 23 2012

seq(n*(n^3-1),n=0..25) ; # R. J. Mathar, Dec 10 2015

Table[n^4 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = { n^4-n } \\ Harry J. Smith, Jun 23 2009

(2*x^2*(7+4*x+x^2)/(1-x)^5).series(x, 37).coefficients(x, sparse=False) # Stefano Spezia, Jul 09 2021

a(n) = n*(n-1)*(n^2+n+1) = A000583(n) - n = A002061(n+1) * A002378(n-1) = (n-1) * A027444(n) = -n * A024001(n).
a(n) = 2*A027482(n). - Zerinvary Lajos, Jan 28 2008
a(n) = floor(n^7/(n^3+1)). - Gary Detlefs, Feb 11 2010
a(n)^3 = (a(n)/n)^4 + (a(n)/n)^3. - Vincenzo Librandi, Feb 23 2012
a(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3, for n > 0. - Vincenzo Librandi, Mar 13 2012
G.f.: 2*x^2*(7 + 4*x + x^2)/(1 - x)^5. - Colin Barker, Apr 23 2012
a(n) = 14*C(n,2) + 36*C(n,3) + 24*C(n,4). - Dennis P. Walsh, Mar 21 2013
Sum_{n>=2} (-1)^n/a(n) = (Pi/3)*sech(Pi*sqrt(3)/2) + 4*log(2)/3 - 1 = 0.06147271494... . - Amiram Eldar, Jul 04 2020
Sum_{n>=2} 1/a(n) = A339605. - R. J. Mathar, Jan 08 2021
E.g.f.: exp(x)*x^2*(7 + 6*x + x^2). - Stefano Spezia, Jul 09 2021
a(n) = 12*A000332(n+2) + 2*A000537(n-1). - Yasser Arath Chavez Reyes, Apr 05 2024

A162614 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^3 - 1.

Original entry on oeis.org

0, 1, 1, 2, 9, 16, 3, 29, 55, 81, 4, 67, 130, 193, 256, 5, 129, 253, 377, 501, 625, 6, 221, 436, 651, 866, 1081, 1296, 7, 349, 691, 1033, 1375, 1717, 2059, 2401, 8, 519, 1030, 1541, 2052, 2563, 3074, 3585, 4096, 9, 737, 1465, 2193, 2921, 3649, 4377, 5105, 5833

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the fourth power of n, A000583(n).

See also the triangles of A162615 and A162616.

			Triangle begins:
  0;
  1,   1;
  2,   9,  16;
  3,  29,  55,  81;
  4,  67, 130, 193, 256;
  5, 129, 253, 377, 501,  625;
  6, 221, 436, 651, 866, 1081, 1296;
  ...

Cf. A000583, A068601, A159797, A162609, A162610, A162611, A162612, A162613, A162615, A162616, A162622, A162623, A162624.

def A162614(n,k):
    return n+k*(n**3-1)
print([A162614(n,k) for n in range(20) for k in range(n+1)])
# R. J. Mathar, Oct 20 2009

Sum_{k=0..n} T(n,k) = n*(n^2-n+1)*(n+1)^2/2 (row sums). - R. J. Mathar, Jul 20 2009

T(n,k) = n + k*(n^3-1). - R. J. Mathar, Oct 20 2009

More terms from R. J. Mathar, Oct 20 2009

cp's OEIS Frontend

A046042 Number of partitions of n into fourth powers.

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A262954 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 has the form z^2 or 8*z^2.

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A209593 T(n,k)=Number of n X n 0..k arrays with every element equal to a diagonal or antidiagonal reflection.

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A262944 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a pentagonal number.

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A262945 Number of ordered pairs (x,y) with x >= 0 and y >= 0 such that n - x^4 - 2*y^2 is a triangular number or a pentagonal number.

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A262956 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a square minus 1.

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A266152 Least positive integer y such that n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists.

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A010801 13th powers: a(n) = n^13.

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A262954 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y(y+1)/2 has the form z^2 or 8z^2.