A000583 -id:A000583 - cp's OEIS frontend

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

1, 4, 15, 32, 65, 108, 175, 256, 369, 500, 671, 864, 1105, 1372, 1695, 2048, 2465, 2916, 3439, 4000, 4641, 5324, 6095, 6912, 7825, 8788, 9855, 10976, 12209, 13500, 14911, 16384, 17985, 19652, 21455, 23328, 25345, 27436, 29679, 32000, 34481, 37044, 39775, 42592

Offset: 1

Stefano Spezia, Aug 01 2018

Terms are obtained as partial sums in an algorithm for the generation of the sequence of the fourth powers (A000583). Starting with the sequence of the positive integers (A000027), it is necessary to delete every 4th term and to consider the partial sums of the obtained sequence, then to delete every 3rd term, and lastly to consider again the partial sums (see References).
a(n) is the trace of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern as shown in the examples below. Specifically, M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even, and it has det(M(n)) = 0 for n > 2 (proved).
From Saeed Barari, Oct 31 2021: (Start)
Also the sum of the entries in an n X n matrix whose elements start from 1 and increase as they approach the center. For instance, in case of n=5, the entries of the following matrix sum to 65:
  1 2 3 2 1
  2 3 4 3 2
  3 4 5 4 3
  2 3 4 3 2
  1 2 3 2 1. (End)
The n X n square matrix of the preceding comment is defined as: A[i,j,n] = n - abs((n + 1)/2 - j) - abs((n + 1)/2 - i). - Stefano Spezia, Nov 05 2021

			For n = 1 the matrix M(1) is
  1
with trace Tr(M(1)) = a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with Tr(M(2)) = a(2) = 4.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with Tr(M(3)) = a(3) = 15.

Edward A. Ashcroft, Anthony A. Faustini, Rangaswami Jagannathan, and William W. Wadge, Multidimensional Programming, Oxford University Press 1995, p. 12.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
G. Polya, Mathematics and Plausible Reasoning: Induction and analogy in mathematics, Princeton University Press 1990, p. 118.
Shailesh Shirali, A Primer on Number Sequences, Universities Press (India) 2004, p. 106.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.7.3 on pages 122-123.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, München 1952. Sitzungsberichte: 1951,3.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A000583, A000027, A186424 (first differences).
Cf. A000578, A000035, A193356, A210378.
Cf. A006003, A317297, A001489, A322844.
Cf. related to the M matrices: A074147 (antidiagonals), A130130 (rank), A241016 (row sums), A317617 (column sums), A322277 (permanent), A323723 (subdiagonal sums), A323724 (superdiagonal sums).

a_n:=List([1..nmax], n->(1/2)*(n^3 + n*RemInt(n, 2)));

List([1..50],n->(1/2)*(n^3+n*(n mod 2))); # Muniru A Asiru, Aug 24 2018

[IsEven(n) select n^3/2 else (n^3+n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 07 2018

a:=n->(1/2)*(n^3+n*modp(n,2)): seq(a(n),n=1..50); # Muniru A Asiru, Aug 24 2018

CoefficientList[Series[1/4 E^-x (1 + 3 E^(2 x) + 6 E^(2 x) x + 2 E^(2 x) x^2), {x, 0, 45}], x]*Table[(k + 1)!, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)*(1 + x)^3), {x, 0, 45}], x]*Table[(k + 1)*(-1)^k, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)^3*(1 + x)), {x, 0, 45}], x]*Table[(k + 1), {k, 0, 45}]
From Robert G. Wilson v, Aug 01 2018: (Start)
a[i_, j_, n_] := If[OddQ@ i, j + n (i - 1), n*i - j + 1]; f[n_] := Tr[Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 45]
CoefficientList[Series[(x^4 + 2x^3 + 6x^2 + 2x + 1)/((x - 1)^4 (x + 1)^2), {x, 0, 45}], x]
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 4, 15, 32, 65, 108}, 45]
(End)

a(n):=(1/2)*(n^3 + n*mod(n,2))$ makelist(a(n), n, 1, nmax);

Vec(x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2) + O(x^40)) \\ Colin Barker, Aug 02 2018

M(i, j, n) = if (i % 2, j + n*(i-1), n*i - j + 1);
a(n) = sum(k=1, n, M(k, k, n)); \\ Michel Marcus, Aug 07 2018

for (n in 1:nmax){
   a <- (n^3+n*n%%2)/2
   output <- c(n, a)
   cat(output, "\n")
}
(MATLAB and FreeMat)
for(n=1:nmax); a=(n^3+n*mod(n,2))/2; fprintf('%d\t%0.f\n',n,a); end

a(n) = (1/2)*(A000578(n) + n*A000035(n)).
a(n) = A006003(n) - (n/2)*(1 - (n mod 2)).
a(n) = Sum_{k=1..n} T(n,k), where T(n,k) = ((n + 1)*k - n)*(n mod 2) + ((n - 1)*k + 1)*(1 - (n mod 2)).
E.g.f.: E(x) = (1/4)*exp(-x)*x*(1 + 3*exp(2*x) + 6*exp(2*x)*x + 2*exp(2*x)*x^2).
L.g.f.: L(x) = -x*(1 + x^2)/((-1 + x)*(1 + x)^3).
H.l.g.f.: LH(x) = -x*(1 + x^2)/((-1 + x)^3*(1 + x)).
Dirichlet g.f.: (1/2)*(Zeta(-3 + s) + 2^(-s)*(-2 + 2^s)*Zeta(-1 + s)).
From Colin Barker, Aug 02 2018: (Start)
G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = n^3/2 for n even.
a(n) = (n^3+n)/2 for n odd. (End)
a(2*n) = A317297(n+1) + A001489(n). - Stefano Spezia, Dec 28 2018
Sum_{n>0} 1/a(n) = (1/2)*(-2*polygamma(0, 1/2) + polygamma(0, (1-i)/2)+ polygamma(0, (1+i)/2)) + zeta(3)/4 approximately equal to 1.3959168891658447368440622669882813003351669... - Stefano Spezia, Feb 11 2019
a(n) = (A000578(n) + A193356(n))/2. - Stefano Spezia, Jun 27 2022
a(n) = A210378(n-1)/n. - Stefano Spezia, Jul 15 2024

A002645 Quartan primes: primes of the form x^4 + y^4, x > 0, y > 0.

Original entry on oeis.org

2, 17, 97, 257, 337, 641, 881, 1297, 2417, 2657, 3697, 4177, 4721, 6577, 10657, 12401, 14657, 14897, 15937, 16561, 28817, 38561, 39041, 49297, 54721, 65537, 65617, 66161, 66977, 80177, 83537, 83777, 89041, 105601, 107377, 119617, 121937

Offset: 1

N. J. A. Sloane

The largest known quartan prime is currently the largest known generalized Fermat prime: The 1353265-digit 145310^262144 + 1 = (145310^65536)^4 + 1^4, found by Ricky L Hubbard. - Jens Kruse Andersen, Mar 20 2011

Primes of the form (a^2 + b^2)/2 such that |a^2 - b^2| is a square. - Thomas Ordowski, Feb 22 2017

			a(1) =   2 = 1^4 + 1^4.
a(2) =  17 = 1^4 + 2^4.
a(3) =  97 = 2^4 + 3^4.
a(4) = 257 = 1^4 + 4^4.

A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929; see Vol. 1, pp. 245-259.
N. D. Elkies, Primes of the form a^4 + b^4, Mathematical Buds, Ed. H. D. Ruderman Vol. 3 Chap. 3 pp. 22-8 Mu Alpha Theta 1984.
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).

Zak Seidov, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
A. J. C. Cunningham, High quartan factorisations and primes, Messenger of Mathematics 36 (1907), pp. 145-174.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. [Annotated scans of a few pages from Volumes 1 and 2]
Ernest G. Hibbs, Component Interactions of the Prime Numbers, Ph. D. Thesis, Capitol Technology Univ. (2022), see p. 33.

Subsequence of A002313 and of A028916.
Intersection of A004831 and A000040.
Cf. A002646, A000583, A003336, A000290, A256852.

a002645 n = a002645_list !! (n-1)
a002645_list = 2 : (map a000040 $ filter ((> 1) . a256852) [1..])
-- Reinhard Zumkeller, Apr 11 2015

nn = 100000; Sort[Reap[Do[n = a^4 + b^4; If[n <= nn && PrimeQ[n], Sow[n]], {a, nn^(1/4)}, {b, a}]][[2, 1]]]
With[{nn=20},Select[Union[Flatten[Table[x^4+y^4,{x,nn},{y,nn}]]],PrimeQ[ #] && #<=nn^4+1&]] (* Harvey P. Dale, Aug 10 2021 *)

upto(lim)=my(v=List(2),t);forstep(x=1,lim^.25,2,forstep(y=2,(lim-x^4)^.25,2,if(isprime(t=x^4+y^4),listput(v,t))));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 05 2011

list(lim)=my(v=List([2]),x4,t); for(x=1,sqrtnint(lim\=1,4), x4=x^4; forstep(y=1+x%2,min(sqrtnint(lim-x4,4), x-1),2, if(isprime(t=x4+y^4), listput(v,t)))); Set(v) \\ Charles R Greathouse IV, Aug 20 2017

A000040 INTERSECTION A003336. - Jonathan Vos Post, Sep 23 2006

A256852(A049084(a(n))) > 1 for n > 1. - Reinhard Zumkeller, Apr 11 2015

More terms from Victoria A Sapko (vsapko(AT)canes.gsw.edu), Nov 07 2002

A270566 Number of ordered ways to write n as x^4 + y(3y+1)/2 + z(7z+1)/2, where x, y and z are integers with x nonnegative.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 19 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 24, 47, 63, 78, 79, 143, 153, 325, 494, 949, 1079, 3328, 4335, 5609, 7949, 7967, 8888, 9665.

Conjecture verified for n up to 10^11. - Mauro Fiorentini, Jul 24 2023

			a(24) = 1 since 24 = 2^4 + (-2)*(3*(-2)+1)/2 + (-1)*(7*(-1)+1)/2.
a(78) = 1 since 78 = 1^4 + 7*(3*7+1)/2 + 0*(7*0+1)/2.
a(143) = 1 since 143 = 1^4 + 6*(3*6+1)/2 + (-5)*(7*(-5)+1)/2.
a(494) = 1 since 494 = 4^4 + (-7)*(3*(-7)+1)/2 + (-7)*(7*(-7)+1)/2.
a(949) = 1 since 949 = 4^4 + 0*(3*0+1)/2 + 14*(7*14+1)/2.
a(1079) = 1 since 1079 = 0^4 + 25*(3*25+1)/2 + 6*(7*6+1)/2.
a(3328) = 1 since 3328 = 0^4 + 38*(3*38+1)/2 + 18*(7*18+1)/2.
a(4335) = 1 since 4335 = 2^4 + 49*(3*49+1)/2 + 14*(7*14+1)/2.
a(5609) = 1 since 5609 = 0^4 + (-61)*(3*(-61)+1)/2 + 4*(7*4+1)/2.
a(7949) = 1 since 7949 = 3^4 + 43*(3*43+1)/2 + 38*(7*38+1)/2.
a(7967) = 1 since 7967 = 7^4 + (-61)*(3*(-61)+1)/2 + 2*(7*2+1)/2.
a(8888) = 1 since 8888 = 0^4 + (-77)*(3*(-77)+1)/2 + 3*(7*3+1)/2.
a(9665) = 1 since 9665 = 3^4 + 73*(3*73+1)/2 + 21*(7*21+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Cf. A001318, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559.

(* From Zhi-Wei Sun, Start *)
pQ[n_] := pQ[n] = IntegerQ[Sqrt[24 n + 1]];
Do[r = 0; Do[If[pQ[n - x^4 - y (7 y + 1)/2], r = r + 1], {x, 0, n^(1/4)}, {y, -Floor[(Sqrt[56 (n - x^4) + 1] + 1)/14], (Sqrt[56 (n - x^4) + 1] - 1)/14}]; Print[n, " ", r]; Continue, {n, 0, 80}]
(* From Zhi-Wei Sun, End *)
A270566[n_] := Length@Solve[x >= 0 && n == x^4 + y*(3 y + 1)/2 + z*(7 z + 1)/2, {x, y, z}, Integers];
Array[A270566, 25, 0] (* JungHwan Min, Mar 19 2016 *)

A357642 Number of even-length integer compositions of 2n whose half-alternating sum is 0.

Original entry on oeis.org

1, 0, 1, 4, 13, 48, 186, 712, 2717, 10432, 40222, 155384, 601426, 2332640, 9063380, 35269392, 137438685, 536257280, 2094786870, 8191506136, 32063203590, 125613386912, 492516592620, 1932569186288, 7588478653938, 29816630378368, 117226929901676, 461151757861552

Offset: 0

Gus Wiseman, Oct 12 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 a(0) = 1 through a(4) = 13 compositions:
  ()  .  (1111)  (1212)  (1313)
                 (1221)  (1322)
                 (2112)  (1331)
                 (2121)  (2213)
                         (2222)
                         (2231)
                         (3113)
                         (3122)
                         (3131)
                         (111311)
                         (112211)
                         (113111)
                         (11111111)

David A. Corneth, Table of n, a(n) for n = 0..1665

The skew-alternating version appears to be A000984.
For original alternating sum we have A001700/A088218.
The version for partitions of any length is A357639, ranked by A357631.
For length multiple of 4 we have A110145.
These compositions of any length are ranked by A357625, reverse A357626.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357621 = half-alternating sum of standard compositions, reverse A357622.
A357637 counts partitions by half-alternating sum, skew A357638.
Cf. A000583, A001511, A035363, A053251, A344619, A357136, A357182, A357627, A357628, A357629, A357633.

Table[Length[Select[Join @@ Permutations/@IntegerPartitions[2n],EvenQ[Length[#]]&&halfats[#]==0&]],{n,0,9}]

a(n) = {my(v, res); if(n < 3, return(1 - bitand(n,1))); res = 0; v = vector(2*n, i, binomial(n-1,i-1)); forstep(i = 4, 2*n, 2, lp = i\4 * 2; rp = i - lp; res += v[lp] * v[rp]; ); res } \\ David A. Corneth, Oct 13 2022

More terms from Alois P. Heinz, Oct 12 2022

A368714 Numbers whose maximal exponent in their prime factorization is even.

Original entry on oeis.org

1, 4, 9, 12, 16, 18, 20, 25, 28, 36, 44, 45, 48, 49, 50, 52, 60, 63, 64, 68, 75, 76, 80, 81, 84, 90, 92, 98, 99, 100, 112, 116, 117, 121, 124, 126, 132, 140, 144, 147, 148, 150, 153, 156, 162, 164, 169, 171, 172, 175, 176, 180, 188, 192, 196, 198, 204, 207, 208

Offset: 1

Amiram Eldar, Jan 04 2024

First differs from A240112 at n = 30.
Numbers k such that A051903(k) is even.
The asymptotic density of this sequence is Sum_{k>=2} (-1)^k * (1 - 1/zeta(k)) = 0.27591672059822700769... .

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A051903, A240112.
Subsequences: A000290, A000302, A000583, A001014, A001016, A001025, A008454, A062503, A067259.
Similar sequences: A060476, A074661, A096432, A336064, A368715.

Select[Range[210], # == 1 || EvenQ[Max[FactorInteger[#][[;;, 2]]]] &]

lista(kmax) = for(k = 1, kmax, if(k == 1 || !(vecmax(factor(k)[,2])%2), print1(k, ", ")));

A005065 Sum of 4th powers of primes dividing n.

Original entry on oeis.org

0, 16, 81, 16, 625, 97, 2401, 16, 81, 641, 14641, 97, 28561, 2417, 706, 16, 83521, 97, 130321, 641, 2482, 14657, 279841, 97, 625, 28577, 81, 2417, 707281, 722, 923521, 16, 14722, 83537, 3026, 97, 1874161, 130337, 28642, 641, 2825761, 2498, 3418801, 14657, 706, 279857, 4879681, 97, 2401, 641, 83602, 28577, 7890481, 97

Offset: 1

N. J. A. Sloane

nonn

Primes are taken without multiplicity, e.g., 12 = 2*2*3, and a(12) = 2^4+3^4 = 97. - Harvey P. Dale, Jul 16 2014

Inverse Möbius transform of n^4 * c(n), where c(n) is the prime characteristic (A010051). - Wesley Ivan Hurt, Jun 22 2024

Antti Karttunen, Table of n, a(n) for n = 1..10000

Column k=4 of A322080.
Cf. A000583, A005068, A005073, A005077, A005081, A005085, A008472, A059841, A079978.
Sum of the k-th powers of the primes dividing n for k=0..10 : A001221 (k=0), A008472 (k=1), A005063 (k=2), A005064 (k=3), this sequence (k=4), A351193 (k=5), A351194 (k=6), A351195 (k=7), this sequence (k=8), A351197 (k=9), A351198 (k=10).
Cf. A010051.

A005065 := proc(n)
        add(d^4, d= numtheory[factorset](n)) ;
end proc;
seq(A005065(n),n=1..40) ; # R. J. Mathar, Nov 08 2011

Join[{0},Table[Total[Transpose[FactorInteger[n]][[1]]^4],{n,2,40}]] (* Harvey P. Dale, Jul 16 2014 *)
Array[DivisorSum[#, #^4 &, PrimeQ] &, 54] (* Michael De Vlieger, Jul 11 2017 *)

a(n) = my(f=factor(n)); sum(k=1, #f~, f[k,1]^4); \\ Michel Marcus, Jul 11 2017

from sympy import primefactors
def a(n): return sum(p**4 for p in primefactors(n))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 11 2017

(define (A005065 n) (if (= 1 n) 0 (+ (A000583 (A020639 n)) (A005065 (A028234 n))))) ;; Antti Karttunen, Jul 10 2017

Additive with a(p^e) = p^4.
From Antti Karttunen, Jul 11 2017: (Start)
a(n) = A005068(n) + 16*A059841(n).
a(n) = A005081(n) + A005085(n) + 16*A059841(n).
a(n) = A005073(n) + A005077(n) + 81*A079978(n).
(End)
G.f.: Sum_{k>=1} prime(k)^4*x^prime(k)/(1 - x^prime(k)). - Ilya Gutkovskiy, Dec 24 2018
a(n) = Sum_{p|n, p prime} p^4. - Wesley Ivan Hurt, Feb 04 2022
a(n) = Sum_{d|n} d^4 * c(d), where c = A010051. - Wesley Ivan Hurt, Jun 22 2024

More terms from Antti Karttunen, Jul 10 2017

A062392 a(n) = n^4 - (n-1)^4 + (n-2)^4 - ... 0^4.

Original entry on oeis.org

0, 1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945, 28680, 36856, 46665, 58311, 72010, 87990, 106491, 127765, 152076, 179700, 210925, 246051, 285390, 329266, 378015, 431985, 491536, 557040, 628881, 707455, 793170, 886446, 987715

Offset: 0

Henry Bottomley, Jun 21 2001

Number of edges in the join of two complete graphs of order n^2 and n, K_n^2 * K_n. - Roberto E. Martinez II, Jan 07 2002
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-5)(P(4,1)-(-1)^k P(4,2k+1))|. - Peter Luschny, Jul 12 2009
Define an infinite symmetric array by T(n,m) = n*(n-1) + m for 0 <= m <= n and T(n,m) = T(m,n), n >= 0. Then a(n) is the sum of terms in the top left (n+1) X (n+1) subarray: a(n) = Sum_{r=0..n} Sum_{c=0..n} T(r,c). - J. M. Bergot, Jul 05 2013
a(n) is the sum of all positive numbers less than A002378(n). - J. M. Bergot, Aug 30 2013
Except the first term, these are triangular numbers that remain triangular when divided by their index, e.g., 66 divided by 11 gives 6. - Waldemar Puszkarz, Sep 14 2017
a(n) is the semiperimeter of the unique primitive Pythagorean triple such that (a-b+c)/2 = T(n) = A000217(n). Its long leg and hypotenuse are consecutive natural numbers and the triple is (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

			From _Bruno Berselli_, Oct 30 2017: (Start)
After 0:
1   =                 -(1) + (2);
15  =             -(1 + 2) + (3 + 4 + 5 + 2*3);
66  =         -(1 + 2 + 3) + (4 + 5 + 6 + 7 + ... + 11 + 3*4);
190 =     -(1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + ... + 19 + 4*5);
435 = -(1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + ... + 29 + 5*6), etc. (End)

T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000538, A000583. A062393 provides the result for 5th powers, A011934 for cubes, A000217 for squares, A001057 (unsigned) for nonnegative integers, A000035 (offset) for 0th powers.
Cf. A000217, A000384, A007588.
Cf. A236770 (see crossrefs).
Cf. A384329, A384369.

a := n -> (2*n^2+n^3-1)*n/2; # Peter Luschny, Jul 12 2009

Table[n (n + 1) (n^2 + n - 1)/2, {n, 0, 40}] (* Harvey P. Dale, Oct 19 2011 *)

{ a=0; for (n=0, 1000, write("b062392.txt", n, " ", a=n^4 - a) ) } \\ Harry J. Smith, Aug 07 2009

a(n) = n*(n+1)*(n^2 + n - 1)/2 = n^4 - a(n-1) = A000583(n) - a(n-1) = A000217(A028387(n-1)) = A000217(n)*A028387(n-1).
a(n) = Sum_{i=0..n} A007588(i) for n > 0. - Jonathan Vos Post, Mar 15 2006
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Harvey P. Dale, Oct 19 2011
G.f.: x*(x*(x + 10) + 1)/(1 - x)^5. - Harvey P. Dale, Oct 19 2011
a(n) = A000384(A000217(n)). - Bruno Berselli, Jan 31 2014
a(n) = A110450(n) - A002378(n). - Gionata Neri, May 13 2015
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*2*Pi/sqrt(5). - Amiram Eldar, Jan 22 2024
a(n) = sqrt(144*A288876(n-2) + 72*A006542(n+2) + A000537(n)). - Yasser Arath Chavez Reyes, Jul 22 2024
E.g.f.: exp(x)*x*(2 + 13*x + 8*x^2 + x^3)/2. - Stefano Spezia, Apr 27 2025
a(n) = A000217(n)*(2*A000217(n)-1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

A113849 Numbers whose prime factors are raised to the fourth power.

Original entry on oeis.org

16, 81, 625, 1296, 2401, 10000, 14641, 28561, 38416, 50625, 83521, 130321, 194481, 234256, 279841, 456976, 707281, 810000, 923521, 1185921, 1336336, 1500625, 1874161, 2085136, 2313441, 2825761, 3111696, 3418801, 4477456, 4879681, 6765201, 7890481

Offset: 1

Cino Hilliard, Jan 25 2006

This is essentially A005117 (the squarefree numbers) raised to the fourth power. - T. D. Noe, Mar 13 2013

All positive integers have a unique factorization into powers of squarefree numbers with distinct exponents that are powers of two. So every positive number is a product of at most one squarefree number (A005117), at most one square of a squarefree number (A062503), at most one 4th power of a squarefree number (term of this sequence), at most one 8th power of a squarefree number, and so on. - Peter Munn, Mar 12 2020

			1296 = 16*81 = 2^4*3^4 so the prime factors of 1296, 2 and 3, are raised to the fourth power.

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

Proper subset of A000583.
Other powers of squarefree numbers: A005117(1), A062503(2), A062838(3),  A113850(5), A113851(6), A113852(7), A072774(all).
Cf. A000351, A225546.

Select[ Range@50^4, Union[Last /@ FactorInteger@# ] == {4} &] (* Robert G. Wilson v, Jan 26 2006 *)
nn = 50; t = Select[Range[2, nn], Union[Transpose[FactorInteger[#]][[2]]] == {1} &]; t^4 (* T. D. Noe, Mar 13 2013 *)
Rest[Select[Range[100], SquareFreeQ]^4] (* Vaclav Kotesovec, May 22 2020 *)

allpwrfact(n,p) = { local(x,j,ln,y,flag); for(x=4,n, y=Vec(factor(x)); ln = length(y[1]); flag=0; for(j=1,ln, if(y[2][j]==p,flag++); ); if(flag==ln,print1(x",")); ) } \\ All prime factors are raised to the power p

from math import isqrt
from sympy import mobius
def A113849(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    kmin, kmax = 1,2
    while f(kmax) >= kmax:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if f(kmid) < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmax**4 # Chai Wah Wu, Aug 19 2024

From Peter Munn, Oct 31 2019: (Start)
a(n) = A005117(n+1)^4.
{a(n)} = {A225546(A000351(n)) : n >= 0} \ {1}, where {a(n)} denotes the set of integers in the sequence.
(End)
Sum_{k>=1} 1/a(k) = zeta(4)/zeta(8) - 1 = 105/Pi^4 - 1. - Amiram Eldar, May 22 2020

More terms from Robert G. Wilson v, Jan 26 2006

A123865 a(n) = n^4 - 1.

Original entry on oeis.org

0, 15, 80, 255, 624, 1295, 2400, 4095, 6560, 9999, 14640, 20735, 28560, 38415, 50624, 65535, 83520, 104975, 130320, 159999, 194480, 234255, 279840, 331775, 390624, 456975, 531440, 614655, 707280, 809999, 923520, 1048575, 1185920, 1336335

Offset: 1

Reinhard Zumkeller, Oct 16 2006

a(n) mod 5 = 0 iff n mod 5 > 0: a(A008587(n)) = 4; a(A047201(n)) = 0; a(n) mod 5 = 4*(1-A079998(n)).

A129292(n) = number of divisors of a(n) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A005563, A024002, A123866, A123867, A123868, A219086.

List([1..40], n-> n^4 -1); # G. C. Greubel, Aug 08 2019

[n^4 - 1: n in [1..40]]; // Vincenzo Librandi, May 01 2011

seq(n^4 -1, n=1..40); # G. C. Greubel, Aug 08 2019

Table[n^4-1,{n,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)

vector(40, n, n^4 -1) \\ G. C. Greubel, Aug 08 2019

[n^4 -1 for n in (1..40)] # G. C. Greubel, Aug 08 2019

G.f.: x^2*(15 + 5*x + 5*x^2 - x^3)/(1-x)^5. - Colin Barker, Jan 10 2012
-4*a(n+1) = -4*n*(n+2)*(n^2+2*n+2) = (n+n*i)*(n+2+n*i)*(n+(n+2)*i)*(n+2+(n+2)*i), where i is the imaginary unit. - Jon Perry, Feb 05 2014
From Vaclav Kotesovec, Feb 14 2015: (Start)
Sum_{n>=2} 1/a(n) = 7/8 - Pi*coth(Pi)/4 = A256919.
Sum_{n>=2} (-1)^n / a(n) = 1/8 - Pi/(4*sinh(Pi)). (End)
a(n) = A005563(A005563(n)). - Bruno Berselli, May 28 2015
E.g.f.: 1 + (-1 + x + 7*x^2 + 6*x^3 + x^4)*exp(x). - G. C. Greubel, Aug 08 2019
Product_{n>=2} (1 + 1/a(n)) = 4*Pi*csch(Pi). - Amiram Eldar, Jan 20 2021

A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 20, 22, 70, 87, 167, 252, 388, 562, 636, 658, 873, 2598, 14979, 18892, 20824.
(ii) Each n = 0,1,2,... can be written as x*P(x) + y^2 + z*(3z-1)/2 with x and y nonnegative integers, and z an integer, where P(x) is either of the polynomials x^3+2, x^3+3, x^3+2x+8, x^3+x^2+4x+2, x^3+x^2+7x+6.
(iii) Any nonnegative integer can be expressed as x*(x^3+3) + y*(5y+4) + z*(3z-1)/2, where x is an nonnegative integer, and y and z are integers.
See also A270516 for a similar conjecture.

			a(20) = 1 since 20 = 1^4 + 1^3 + 4^2 + (-1)*(3*(-1)-1)/2.
a(22) = 1 since 22 = 0^4 + 0^3 + 0^2 + 4*(3*4-1)/2.
a(873) = 1 since 873 = 5^4 + 5^3 + 11^2 + (-1)*(3*(-1)-1)/2.
a(2598) = 1 since 2598 =  4^4 + 4^3 + 4^2 + 39*(3*39-1)/2.
a(14979) = 1 since 14979 = 1^4 + 1^3 + 51^2 + 91*(3*91-1)/2.
a(18892) = 1 since 18892 = 3^4 + 3^3 + 137^2 + (-3)*(3*(-3)-1)/2.
a(20824) = 1 since 20824 = 1^4 + 1^3 + 115^2 + (-71)*(3*(-71)-1)/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. A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516.

pQ[x_]:=pQ[x]=IntegerQ[Sqrt[24x+1]]
Do[r=0;Do[If[pQ[n-y^2-x^3*(x+1)],r=r+1],{y,0,Sqrt[n]},{x,0,(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

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A317614 a(n) = (1/2)*(n^3 + n*(n mod 2)).

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A002645 Quartan primes: primes of the form x^4 + y^4, x > 0, y > 0.

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A270566 Number of ordered ways to write n as x^4 + y*(3y+1)/2 + z*(7z+1)/2, where x, y and z are integers with x nonnegative.

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A357642 Number of even-length integer compositions of 2n whose half-alternating sum is 0.

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A368714 Numbers whose maximal exponent in their prime factorization is even.

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A005065 Sum of 4th powers of primes dividing n.

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A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

A270566 Number of ordered ways to write n as x^4 + y(3y+1)/2 + z(7z+1)/2, where x, y and z are integers with x nonnegative.