A000583 -id:A000583 - cp's OEIS frontend

A162622 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^4 - 1.

0, 1, 1, 2, 17, 32, 3, 83, 163, 243, 4, 259, 514, 769, 1024, 5, 629, 1253, 1877, 2501, 3125, 6, 1301, 2596, 3891, 5186, 6481, 7776, 7, 2407, 4807, 7207, 9607, 12007, 14407, 16807, 8, 4103, 8198, 12293, 16388, 20483, 24578, 28673, 32768, 9, 6569, 13129

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the 5th power of n, A000584(n).

See also the triangles of A162623 and A162624.

			Triangle begins:
  0;
  1,    1;
  2,   17,    32;
  3,   83,   163,   243;
  4,  259,   514,   769,  1024;
  5,  629,  1253,  1877,  2501,  3125;
  6, 1301,  2596,  3891,  5186,  6481,  7776;
  7, 2407,  4807,  7207,  9607, 12007, 14407, 16807;
  8, 4103,  8198, 12293, 16388, 20483, 24578, 28673, 32768;
  9, 6569, 13129, 19689, 26249, 32809, 39369, 45929, 52489, 59049; etc.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A000583, A000584, A123865, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162615, A162616, A162623, A162624.

/* Triangle: */ [[n+k*(n^4-1): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Dec 14 2012

A162622 := proc(n,k) n+k*(n^4-1) ; end proc: seq(seq( A162622(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Feb 11 2010

Flatten[Table[NestList[#+n^4-1&,n,n],{n,0,9}]] (* Harvey P. Dale, Jun 23 2013 *)

Sum_{k=0..n} T(n,k) = n*(n+1)*(1+n^4)/2 (row sums). [R. J. Mathar, Jul 20 2009]

7th and later rows from R. J. Mathar, Feb 11 2010

A252895 Numbers with an odd number of square divisors.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 26, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 48, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 96, 97

Offset: 1

Walker Dewey Anderson, Mar 22 2015

nonn

Open lockers in the locker problem where the student numbers are the set of perfect squares.
The locker problem is a classic mathematical problem. Imagine a row containing an infinite number of lockers numbered from one to infinity. Also imagine an infinite number of students numbered from one to infinity. All of the lockers begin closed. The first student opens every locker that is a multiple of one, which is every locker. The second student closes every locker that is a multiple of two, so all of the even-numbered lockers are closed. The third student opens or closes every locker that is a multiple of three. This process continues for all of the students. [This is sometimes called the light switch problem - see A360845.]
A variant on the locker problem is when not all student numbers are considered; in the case of this sequence, only the square-numbered students open and close lockers. The sequence here is a list of the open lockers after all of the students have gone.
n is in the sequence if and only if it is the product of a squarefree number (A005117) and a fourth power (A000583). - Robert Israel, Apr 07 2015
Let D be the multiset containing d0(k), the divisor counting function, for each divisor k of n. n is in the sequence if and only if D admits a partition into two parts A and B such that the sum of the elements of A is exactly one more or less than the sum of the elements of B. For example, if n = 80, we have D = {1, 2, 2, 3, 4, 4, 5, 6, 8, 10}, and A = {1, 2, 3, 4, 4, 8} and B = {2, 5, 6, 10}. The sum of A is 22, and the sum of B is 23. - Griffin N. Macris, Oct 10 2016
From Amiram Eldar, Jul 07 2020: (Start)
Numbers k such that the largest square dividing k (A008833) is a fourth power.
The asymptotic density of this sequence is Pi^2/15 = A182448 = 0.657973... (Cesàro, 1885). (End)
Closed under the binary operation A059897(.,.), forming a subgroup of the positive integers under A059897. - Peter Munn, Aug 01 2020

			The set of divisors of 6 is {1,2,3,6}, which contains only one perfect square: 1; therefore 6 is a term.
The set of divisors of 16 is {1,2,4,8,16}, which contains three perfect squares: 1, 4, and 16; therefore 16 is a term.
The set of divisors of 4 is {1,2,4}, which contains two perfect squares: 1 and 4; therefore 4 is not a term.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Ernest Cesàro, Le plus grand diviseur carré, Annali di Matematica Pura ed Applicata, Vol. 13, No. 1 (1885), pp. 251-268, entire volume.
K. A. P. Dagal, Generalized Locker Problem, arXiv:1307.6455 [math.NT], 2013.
B. Torrence and S. Wagon, The Locker Problem, Crux Mathematicorum, 2007, 33(4), 232-236.

Cf. A000290, A000583, A005117, A008833, A046951, A182448, A252849, A360845.

Positions of ones in A335324.

a252895 n = a252895_list !! (n-1)
a252895_list = filter (odd . a046951) [1..]
-- Reinhard Zumkeller, Apr 06 2015

N:= 1000: # to get all terms <= N
S:= select(numtheory:-issqrfree, {$1..N}):
map(s -> seq(s*i^4, i = 1 .. floor((N/s)^(1/4))), S);
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(%,list)); # Robert Israel, Apr 07 2015

Position[Length@ Select[Divisors@ #, IntegerQ@ Sqrt@ # &] & /@ Range@ 70, Integer?OddQ] // Flatten (* _Michael De Vlieger, Mar 23 2015 *)
a[n_] := DivisorSigma[0, Total[EulerPhi/@Select[Sqrt[Divisors[n]], IntegerQ]]]; Flatten[Position[a/@Range@100,?OddQ]] (* _Ivan N. Ianakiev, Apr 07 2015 *)
Select[Range@ 100, OddQ@ Length@ DeleteCases[Divisors@ #, k_ /; ! IntegerQ@ Sqrt@ k] &] (* Michael De Vlieger, Oct 10 2016 *)

isok(n) = sumdiv(n, d, issquare(d)) % 2; \\ Michel Marcus, Mar 22 2015

[n for n in [1..200] if len([x for x in divisors(n) if is_square(x)])%2==1] # Tom Edgar, Mar 22 2015

A284900 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^4.

Original entry on oeis.org

1, 15, 82, 239, 626, 1230, 2402, 3823, 6643, 9390, 14642, 19598, 28562, 36030, 51332, 61167, 83522, 99645, 130322, 149614, 196964, 219630, 279842, 313486, 391251, 428430, 538084, 574078, 707282, 769980, 923522, 978671, 1200644, 1252830, 1503652, 1587677

Offset: 1

Seiichi Manyama, Apr 05 2017

Multiplicative because this sequence is the Dirichlet convolution of A000583 and A062157 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

Seiichi Manyama, Table of n, a(n) for n = 1..10000
J. W. L. Glaisher, On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
Index entries for sequences mentioned by Glaisher.

Sum_{d|n} (-1)^(n/d+1)*d^k: A000593 (k=1), A078306 (k=2), A078307 (k=3), this sequence (k=4), A284926 (k=5), A284927 (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Cf. A000583, A013663, A062157, A386729.

Table[Sum[(-1)^(n/d + 1)*d^4, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 05 2017 *)
f[p_, e_] := (p^(4*e + 4) - 1)/(p^4 - 1); f[2, e_] := (7*2^(4*e + 1) + 1)/15; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)

a(n) = sumdiv(n, d, (-1)^(n/d + 1)*d^4); \\ Indranil Ghosh, Apr 05 2017

from sympy import divisors
print([sum([(-1)**(n//d + 1)*d**4 for d in divisors(n)]) for n in range(1, 51)]) # Indranil Ghosh, Apr 05 2017

G.f.: Sum_{k>=1} k^4*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (7*2^(4*e+1)+1)/15, and a(p^e) = (p^(4*e+4) - 1)/(p^4 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^5, where c = 3*zeta(5)/16 = 0.194423... . (End)

A357635 Numbers k such that the half-alternating sum of the partition having Heinz number k is 1.

Original entry on oeis.org

2, 8, 24, 32, 54, 128, 135, 162, 375, 384, 512, 648, 864, 875, 1250, 1715, 1944, 2048, 2160, 2592, 3773, 4374, 4802, 5000, 6000, 6144, 8192, 9317, 10368, 10935, 13122, 13824, 14000, 15000, 17303, 19208, 20000, 24167, 27440, 29282, 30375, 31104, 32768, 33750

Offset: 1

Gus Wiseman, Oct 28 2022

nonn

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

			The terms together with their prime indices begin:
    2: {1}
    8: {1,1,1}
   24: {1,1,1,2}
   32: {1,1,1,1,1}
   54: {1,2,2,2}
  128: {1,1,1,1,1,1,1}
  135: {2,2,2,3}
  162: {1,2,2,2,2}
  375: {2,3,3,3}
  384: {1,1,1,1,1,1,1,2}
  512: {1,1,1,1,1,1,1,1,1}
  648: {1,1,1,2,2,2,2}
  864: {1,1,1,1,1,2,2,2}
  875: {3,3,3,4}

The version for k = 0 is A000583, standard compositions A357625-A357626.
The version for original alternating sum is A345958.
Positions of ones in A357633, non-reverse A357629.
The skew version for k = 0 is A357636, non-reverse A357632.
These partitions are counted by A035444, skew A035544.
The non-reverse version is A357851, k = 0 version A357631.
A056239 adds up prime indices, row sums of A112798.
A316524 gives alternating sum of prime indices, reverse A344616.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357641 counts comps w/ half-alt sum 0, even-length A357642.
Cf. A000290, A003963, A053251, A055932, A357621-A357624, A357630, A357634, A357637, A357639, A357640.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
Select[Range[1000],halfats[Reverse[primeMS[#]]]==1&]

A055565 Sum of digits of n^4.

Original entry on oeis.org

0, 1, 7, 9, 13, 13, 18, 7, 19, 18, 1, 16, 18, 22, 22, 18, 25, 19, 27, 10, 7, 27, 22, 31, 27, 25, 37, 18, 28, 25, 9, 22, 31, 27, 25, 19, 36, 28, 25, 18, 13, 31, 27, 25, 37, 18, 37, 43, 27, 31, 13, 27, 25, 37, 27, 28, 43, 18, 31, 22, 18, 34, 37, 36, 37, 34, 45, 13, 31, 27, 7

Offset: 0

Henry Bottomley, Jun 19 2000

			a(2) = 7 because 2^4 = 16 and 1+6 = 7.

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A000583, A007953, A055570, A055575 (fixed points), A373914.

Other powers: A004159, A004164, A055566, A055567, A066588.

for i from 0 to 200 do printf(`%d,`,add(j, j=convert(i^4, base, 10))) od;

a[n_Integer]:=Apply[Plus, IntegerDigits[n^4]]; Table[a[n], {n, 0, 100}] (* Vincenzo Librandi, Feb 23 2015 *)

a(n) = sumdigits(n^4); \\ Seiichi Manyama, Nov 16 2021

[sum((n^4).digits()) for n in (0..70)] # Bruno Berselli, Feb 23 2015

a(n) = A007953(A000583(n)). - Michel Marcus, Feb 23 2015

More terms from James Sellers, Jul 04 2000

A082043 Square array, A(n, k) = (kn)^2 + 2k*n + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 25, 16, 1, 1, 25, 49, 49, 25, 1, 1, 36, 81, 100, 81, 36, 1, 1, 49, 121, 169, 169, 121, 49, 1, 1, 64, 169, 256, 289, 256, 169, 64, 1, 1, 81, 225, 361, 441, 441, 361, 225, 81, 1, 1, 100, 289, 484, 625, 676, 625, 484, 289, 100, 1

Offset: 0

Paul Barry, Apr 03 2003

			Array, A(n, k), begins as:
  1,   1,   1,    1,    1,    1,    1,    1,     1, ... A000012;
  1,   4,   9,   16,   25,   36,   49,   64,    81, ... A000290;
  1,   9,  25,   49,   81,  121,  169,  225,   289, ... A016754;
  1,  16,  49,  100,  169,  256,  361,  484,   625, ... A016778;
  1,  25,  81,  169,  289,  441,  625,  841,  1089, ... A016814;
  1,  36, 121,  256,  441,  676,  961, 1296,  1681, ... A016862;
  1,  49, 169,  361,  625,  961, 1369, 1849,  2401, ... A016922;
  1,  64, 225,  484,  841, 1296, 1849, 2500,  3249, ... A016994;
  1,  81, 289,  625, 1089, 1681, 2401, 3249,  4225, ... A017078;
  1, 100, 361,  784, 1369, 2116, 3025, 4096,  5329, ... A017174;
  1, 121, 441,  961, 1681, 2601, 3721, 5041,  6561, ... A017282;
  1, 144, 529, 1156, 2025, 3136, 4489, 6084,  7921, ... A017402;
  1, 169, 625, 1369, 2401, 3721, 5329, 7225,  9409, ... A017534;
  1, 196, 729, 1600, 2809, 4356, 6241, 8464, 11025, ... ;
Antidiagonals, T(n, k), begin as:
  1;
  1,   1;
  1,   4,   1;
  1,   9,   9,   1;
  1,  16,  25,  16,   1;
  1,  25,  49,  49,  25,   1;
  1,  36,  81, 100,  81,  36,   1;
  1,  49, 121, 169, 169, 121,  49,   1;
  1,  64, 169, 256, 289, 256, 169,  64,   1;
  1,  81, 225, 361, 441, 441, 361, 225,  81,   1;
  1, 100, 289, 484, 625, 676, 625, 484, 289, 100,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Rows include A000290, A016754, A016778, A016814, A016862, A016922, A016994, A017078, A017174, A017282, A017402, A017534.
Diagonals include A000583, A058031, A062938, A082044 (main diagonal).
Diagonal sums (row sums if viewed as number triangle) are A082045.
Cf. A082039, A082045, A082046, A082105.

A082043:= func< n,k | (k*(n-k))^2 +2*k*(n-k) +1 >;
[A082043(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 24 2022

T[n_, k_]:= (k*(n-k))^2 +2*k*(n-k) +1;
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 24 2022 *)

def A082043(n,k): return (k*(n-k))^2 +2*k*(n-k) +1
flatten([[A082043(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Dec 24 2022

A(n, k) = (k*n)^2 + 2*k*n + 1 (square array).
T(n, k) = (k*(n-k))^2 + 2*k*(n-k) + 1 (number triangle).
A(k, n) = A(n, k).
T(n, n-k) = T(n, k).
A(n, n) = T(2*n, n) = A082044(n).
A(n, n-1) = T(2*n+1, n-1) = A058031(n), n >= 1.
A(n, n-2) = T(2*(n-1), n) = A000583(n-1), n >= 2.
A(n, n-3) = T(2*n-3, n) = A062938(n-3), n >= 3.
Sum_{k=0..n} T(n, k) = A082045(n) (diagonal sums).
Sum_{k=0..n} (-1)^k * T(n, k) = (1/4)*(1+(-1)^n)*(2 - 3*n). - G. C. Greubel, Dec 24 2022

A231303 Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 16, 82, 272, 707, 1568, 3108, 5664, 9669, 15664, 24310, 36400, 52871, 74816, 103496, 140352, 187017, 245328, 317338, 405328, 511819, 639584, 791660, 971360, 1182285, 1428336, 1713726, 2042992, 2421007, 2852992, 3344528, 3901568, 4530449, 5237904

Offset: 0

Stanislav Sykora, Nov 07 2013

In physics, a(n)/2^(M-1) is the trace of the spin operator |S_z|^M for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of their 4th powers is 2*82/16 = a(3)/8 (analogously for other values of M).

Partial sums of A062392. - Bruce J. Nicholson, Jun 29 2019

			a(4) = 4^4 + 2^4 = 272; a(5) = 5^4 + 3^4 + 1^4 = 707.

Stanislav Sykora, Table of n, a(n) for n = 0..9999
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A001477 (M=1), A000292 (M=2), A105636 (M=3), A231304 (M=5), A231305 (M=6), A231306 (M=7), A231307 (M=8), A231308 (M=9), A231309 (M=10).

Cf. A000389, A000579, A000332, A000583.

List([0..40], n-> n*(3*n^4+15*n^3+20*n^2-8)/30); # G. C. Greubel, Jul 01 2019

[1/30*n*(3*n^4+15*n^3+20*n^2-8): n in [0..40]]; // Vincenzo Librandi, Dec 23 2015

Table[SeriesCoefficient[x*(1+10*x+x^2)/(1-x)^6, {x, 0, n}], {n, 0, 40}] (* Michael De Vlieger, Dec 22 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 16, 82, 272, 707}, 40] (* Vincenzo Librandi, Dec 23 2015 *)

nmax=40;a = vector(nmax);a[2]=1;for(i=3,#a,a[i]=a[i-2]+(i-1)^4); print(a);

concat(0, Vec(x*(1+10*x+x^2)/(1-x)^6 + O(x^40))) \\ Colin Barker, Dec 22 2015

[n*(3*n^4+15*n^3+20*n^2-8)/30 for n in (0..40)] # G. C. Greubel, Jul 01 2019

a(n) = Sum_{k=0..floor(n/2)} (n - 2*k)^4.
From Colin Barker, Dec 22 2015: (Start)
a(n) = (1/30)*n*(3*n^4 + 15*n^3 + 20*n^2 - 8).
G.f.: x*(1 + 10*x + x^2) / (1-x)^6.
(End)
E.g.f.: x*(30 + 210*x + 185*x^2 + 45*x^3 + 3*x^4)*exp(x)/30. - G. C. Greubel, Apr 24 2016
From Bruce J. Nicholson, Jun 29 2019: (Start)
a(n) = 12*A000389(n+3) + A000292(n);
a(n) = (12*A000579(n+4)+A000332(n+3)) - (12*A000579(n+3)+A000332(n+2));
a(n) - a(n-2) = A000583(n). (End)

A232713 Doubly pentagonal numbers: a(n) = n(3n-2)(3n-1)(3n+1)/8.

Original entry on oeis.org

0, 1, 35, 210, 715, 1820, 3876, 7315, 12650, 20475, 31465, 46376, 66045, 91390, 123410, 163185, 211876, 270725, 341055, 424270, 521855, 635376, 766480, 916895, 1088430, 1282975, 1502501, 1749060, 2024785, 2331890, 2672670, 3049501, 3464840, 3921225, 4421275

Offset: 0

Bruno Berselli, Nov 28 2013

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000326, A000332.
Cf. similar sequences: A000583 for A000290(A000290(n)); A002817 for A000217(A000217(n)); A063249 for A000384(A000384(n)).
Cf. A236770, A260810.

[n*(3*n-2)*(3*n-1)*(3*n+1)/8: n in [0..40]];

Table[n (3 n - 2) (3 n - 1) (3 n + 1)/8, {n, 0, 40}]

a(n)=n*(3*n-2)*(3*n-1)*(3*n+1)/8 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(1 + 30*x + 45*x^2 + 5*x^3) / (1 - x)^5.
a(n) = A000326(A000326(n)) = A000332(3n+1).
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 4 + 2*Pi/sqrt(3) - 6*log(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 32*log(2)/3 - 4*Pi/(3*sqrt(3)) - 4. (End)

A346110 Numbers whose square can be represented in exactly four ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

469625, 1878500, 2224625, 4226625, 7514000, 8898500, 11740625, 15289625, 16906500, 20021625, 23011625, 25716665, 30056000, 35594000, 38039625, 46962500, 54316275, 55615625, 56824625, 61158500, 67626000, 79366625, 80086500, 92046500, 92481870

Offset: 1

Karl-Heinz Hofmann, Jul 05 2021

nonn

Terms are numbers z such that there are exactly four solutions to z^2 = x^2 + y^4, where x, y and z belong to the set of positive integers.
Terms cannot be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
If the terms additionally have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.
The special prime factor 2 has the same behavior. Means: If the term is even, x and y must be even too.
Apparently, all terms are divisible by 65. The divided terms are in A346594. Are there exceptions for n > 25? - Hugo Pfoertner, Jul 14 2021, Jul 29 2021
Yes, there are exceptions: a(44,46,53,95,97) are not divisible by 65 (5*13) but they have in common: They are divisible by 145 (5*29). - Karl-Heinz Hofmann, Aug 28 2021

			29679^2 = 29640^2 + 39^4, so 29679 is not a term (only one solution).
60^2 + 5^4 = 63^2 + 4^4 = 65^2, so 65 is not a term (only two solutions).
572^2 + 39^4 = 1500^2 + 25^4 = 1575^2 + 20^4 = 1625^2, so 1625 is not a term (only three solutions).
165308^2 + 663^4 = 349575^2 + 560^4 = 433500^2 + 425^4 = 455175^2 + 340^4 = 469625^2, so 469625 is a term (four solutions).

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 97 terms from Karl-Heinz Hofmann)
Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2,x3,y3,x4,y4} sets up to 10^9.
Karl-Heinz Hofmann, A 3D Animation of the solutions up to 10^9.
Karl-Heinz Hofmann, Python code (not only for 4 Solutions).

Cf. A000290, A000583.
Cf. A271576 (all solutions), A345645 (one solution), A345700 (two solutions), A345968 (three solutions), A348655 (five solutions), A349324 (six solutions), A346115 (the least solutions).
Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

Python
```
# See Hofmann link.
```

A016792 a(n) = (3*n+2)^4.

Original entry on oeis.org

16, 625, 4096, 14641, 38416, 83521, 160000, 279841, 456976, 707281, 1048576, 1500625, 2085136, 2825761, 3748096, 4879681, 6250000, 7890481, 9834496, 12117361, 14776336, 17850625, 21381376, 25411681, 29986576, 35153041, 40960000, 47458321, 54700816, 62742241, 71639296

Offset: 0

N. J. A. Sloane

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

Cf. A016789, A016790, A016791.

Subsequence of A000583.

[(3*n+2)^4 : n in [0..30]]; // Vincenzo Librandi, Sep 29 2011

Table[(3n+2)^4,{n,0,100}] (* Mohammad K. Azarian, Jun 15 2016 *)
LinearRecurrence[{5,-10,10,-5,1},{16,625,4096,14641,38416},30] (* Harvey P. Dale, Aug 02 2018 *)

From Ilya Gutkovskiy, Jun 16 2016: (Start)
G.f.: (16 + 545*x + 1131*x^2 + 251*x^3 + x^4)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
From Amiram Eldar, Mar 31 2022: (Start)
a(n) = A016789(n)^4 = A016790(n)^2.
Sum_{n>=0} 1/a(n) = PolyGamma(3, 2/3)/486. (End)

cp's OEIS Frontend

A162622 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^4 - 1.

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A252895 Numbers with an odd number of square divisors.

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A284900 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^4.

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A357635 Numbers k such that the half-alternating sum of the partition having Heinz number k is 1.

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A055565 Sum of digits of n^4.

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A082043 Square array, A(n, k) = (k*n)^2 + 2*k*n + 1, read by antidiagonals.

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A231303 Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.

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A082043 Square array, A(n, k) = (kn)^2 + 2k*n + 1, read by antidiagonals.

A232713 Doubly pentagonal numbers: a(n) = n(3n-2)(3n-1)(3n+1)/8.