A000583 -id:A000583 - cp's OEIS frontend

A016948 a(n) = (6*n + 3)^4.

81, 6561, 50625, 194481, 531441, 1185921, 2313441, 4100625, 6765201, 10556001, 15752961, 22667121, 31640625, 43046721, 57289761, 74805201, 96059601, 121550625, 151807041, 187388721, 228886641, 276922881, 332150625, 395254161, 466948881, 547981281, 639128961

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A016756, A016945, A016946, A016947.

[(6*n+3)^4: n in [0..50]]; // Vincenzo Librandi, May 05 2011

a[n_] := (6*n + 3)^4; Array[a, 50, 0] (* Amiram Eldar, Mar 30 2022 *)

From Amiram Eldar, Mar 30 2022: (Start)
a(n) = A016945(n)^4 = A016946(n)^2.
a(n) = 3^4*A016756(n).
Sum_{n>=0} 1/a(n) = Pi^4/7776. (End)

A016960 a(n) = (6*n + 4)^4.

Original entry on oeis.org

256, 10000, 65536, 234256, 614656, 1336336, 2560000, 4477456, 7311616, 11316496, 16777216, 24010000, 33362176, 45212176, 59969536, 78074896, 100000000, 126247696, 157351936, 193877776, 236421376, 285610000, 342102016, 406586896, 479785216, 562448656, 655360000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A016792, A016957, A016958, A016959.

Subsequence of A000583.

[(6*n+4)^4: n in [0..40]]; // Vincenzo Librandi, May 06 2011

(6*Range[0,20]+4)^4 (* or *) LinearRecurrence[{5,-10,10,-5,1},{256,10000,65536,234256,614656},30] (* Harvey P. Dale, Sep 23 2013 *)

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Sep 23 2013
G.f.: 16*(16+545*x+1131*x^2+251*x^3+x^4)/(1-x)^5. - Harvey P. Dale, Aug 21 2021
From Amiram Eldar, Mar 31 2022: (Start)
a(n) = A016957(n)^4 = A016958(n)^2.
a(n) = 16*A016792(n).
Sum_{n>=0} 1/a(n) = PolyGamma(3, 2/3)/7776. (End)

A016972 a(n) = (6*n + 5)^4.

Original entry on oeis.org

625, 14641, 83521, 279841, 707281, 1500625, 2825761, 4879681, 7890481, 12117361, 17850625, 25411681, 35153041, 47458321, 62742241, 81450625, 104060401, 131079601, 163047361, 200533921, 244140625, 294499921, 352275361, 418161601, 492884401, 577200625, 671898241

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 (5,-10,10,-5,1).

Cf. A016969, A016970, A016971.

Subsequence of A000583.

[(6*n+5)^4: n in [0..40]]; // Vincenzo Librandi, May 07 2011

Table[(6 n + 5)^4, {n, 0, 20}] (* Michael De Vlieger, Mar 20 2017 *)

From Chai Wah Wu, Mar 20 2017: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4.
G.f.: (-x^4 - 2396*x^3 - 16566*x^2 - 11516*x - 625)/(x - 1)^5. (End)
From Amiram Eldar, Apr 01 2022: (Start)
a(n) = A016969(n)^4 = A016970(n)^2.
Sum_{n>=0} 1/a(n) = PolyGamma(3, 5/6)/7776. (End)

A024196 a(n) = 2nd elementary symmetric function of the first n+1 odd positive integers.

Original entry on oeis.org

3, 23, 86, 230, 505, 973, 1708, 2796, 4335, 6435, 9218, 12818, 17381, 23065, 30040, 38488, 48603, 60591, 74670, 91070, 110033, 131813, 156676, 184900, 216775, 252603, 292698, 337386, 387005, 441905, 502448, 569008, 641971, 721735, 808710, 903318

Offset: 1

Clark Kimberling

			a(8) = 8*80+7*79+6*78+5*77+4*76+3*75+2*74+1*73 = 2796. - _Bruno Berselli_, Mar 13 2012

Bruno Berselli, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

From Johannes W. Meijer, Jun 08 2009: (Start)
Equals third right hand column of A028338 triangle.
Equals third left hand column of A109692 triangle.
Equals third right hand column of A161198 triangle divided by 2^m.
(End)
Cf. A016061.

List([1..36], n -> n*(n+1)*(3*n^2+5*n+1)/6); # Muniru A Asiru, Feb 13 2018

seq(n*(n+1)*(3*n^2+5*n+1)/6,n=1..25); # Muniru A Asiru, Feb 13 2018

f[k_] := 2 k - 1; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[2, t[n]]
Table[a[n], {n, 2, 50}]  (* A024196 *)
(* Clark Kimberling, Dec 31 2011 *)
Table[(n(n+1)(3n^2+5n+1))/6,{n,50}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{3,23,86,230,505},50] (* Harvey P. Dale, Jul 08 2019 *)

a(n) = n*(n+1)*(3*n^2+5*n+1)/6.
From Bruno Berselli, Mar 13 2012: (Start)
G.f.: x*(3 + 8*x + x^2)/(1 - x)^5.
a(n) = Sum_{i=1..n} (n+1-i)*((n+1)^2-i).
a(n) = n*A016061(n) - Sum_{i=0..n-1} A016061(i). (End)
a(n) - a(n-1) = A099721(n). Partial sums of A099721.- Philippe Deléham, May 07 2012
a(n) = Sum_{i=1..n} ((2*i-1)*Sum_{j=i..n} (2*j+1)) = 1*(3+5+...2*n+1) + 3*(5+7+...+2*n+1) + ... + (2*n-1)*(2*n+1). - J. M. Bergot, Apr 21 2017
a(n) = A028338(n+1, n-1), n >= 1, (third diagonal). See the crossref. below. Wolfdieter Lang, Jul 21 2017
a(n) = (A000583(n+1) - A000447(n+1))/2. - J. M. Bergot, Feb 13 2018

A062075 a(n) = n^4 * 4^n.

Original entry on oeis.org

0, 4, 256, 5184, 65536, 640000, 5308416, 39337984, 268435456, 1719926784, 10485760000, 61408804864, 347892350976, 1916696264704, 10312216477696, 54358179840000, 281474976710656, 1434879854116864, 7213895789838336, 35822363710849024, 175921860444160000, 855336483526017024

Offset: 0

Jason Earls, Jun 27 2001

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (20,-160,640,-1280,1024).

Cf. A000302, A000583, A062074.

A062075:=n->n^4*4^n: seq(A062075(n), n=0..30); # Wesley Ivan Hurt, Apr 09 2017

f[n_]:=n^4*4^n; f[Range[0,40]] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011 *)

PARI
```
a(n) = (n^4)*(4^n)
```

[4^n*n^4 for n in (0..30)] # G. C. Greubel, May 10 2022

G.f.: 4*x*(1+4*x)*(1+40*x+16*x^2) / (1-4*x)^5. - Colin Barker, Apr 30 2013

E.g.f.: 4*x*(1 + 28*x + 96*x^2 + 64*x^3)*exp(4*x). - G. C. Greubel, May 10 2022

A092181 Figurate numbers based on the 24-cell (4-D polytope with Schlaefli symbol {3,4,3}).

Original entry on oeis.org

1, 24, 153, 544, 1425, 3096, 5929, 10368, 16929, 26200, 38841, 55584, 77233, 104664, 138825, 180736, 231489, 292248, 364249, 448800, 547281, 661144, 791913, 941184, 1110625, 1301976, 1517049, 1757728, 2025969, 2323800, 2653321, 3016704

Offset: 1

Michael J. Welch (mjw1(AT)ntlworld.com), Mar 31 2004

This is the 4-dimensional regular convex polytope called the 24-cell, hyperdiamond or icositetrachoron.

			a(3)= 3^2*((3*3^2)-(4*3)+2) = 9*(27-12+2) = 9*17 = 153

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Eric Weisstein's World of Mathematics, 24-Cell
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1). [_R. J. Mathar_, Jun 21 2010]

Cf. A000332, A000583, A014820, A092182, A092183.

[n^2*((3*n^2)-(4*n)+2): n in [1..40]]; // Vincenzo Librandi, May 22 2011

Table[SeriesCoefficient[x (1 + 19 x + 43 x^2 + 9 x^3)/(1 - x)^5, {x, 0, n}], {n, 32}] (* Michael De Vlieger, Dec 14 2015 *)
LinearRecurrence[{5,-10,10,-5,1},{1,24,153,544,1425},40] (* Harvey P. Dale, May 25 2022 *)

a(n) = n^2*(3*n^2-4*n+2); \\ Michel Marcus, Dec 14 2015

a(n) = n^2*((3*n^2)-(4*n)+2).
a(n) = C(n+3,4) + 19 C(n+2,4) + 43 C(n+1,4) + 9 C(n,4).
a(n) = +5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5). G.f.: x*(1+19*x+43*x^2+9*x^3)/(1-x)^5. [R. J. Mathar, Jun 21 2010]
a(n) = Sum_{k = 1..n} (k^3 + k^7)* binomial(n,k)/binomial(n+k,k). Cf. A034262 and A155977. - Peter Bala, Feb 12 2019

A092182 Figurate numbers based on the 600-cell (4-D polytope with Schlaefli symbol {3,3,5}).

Original entry on oeis.org

1, 120, 947, 3652, 9985, 22276, 43435, 76952, 126897, 197920, 295251, 424700, 592657, 806092, 1072555, 1400176, 1797665, 2274312, 2839987, 3505140, 4280801, 5178580, 6210667, 7389832, 8729425, 10243376, 11946195, 13852972, 15979377

Offset: 1

Michael J. Welch (mjw1(AT)ntlworld.com), Mar 31 2004

This is the 4-dimensional regular convex polytope called the 600-cell, hexacosichoron or hypericosahedron.

			a(3)= 3*((145*3^3)-(280*3^2)+(179*3)-38)/6 = 3*(3915-2520+537-38)/6 = 0.5*1894 = 947

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric Weisstein's World of Mathematics, 600-Cell
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1). [_R. J. Mathar_, Jun 21 2010]

Cf. A000332, A000583, A014820, A092181, A092183.

[n*((145*n^3)-(280*n^2)+(179*n)-38)/6: n in [1..40]]; // Vincenzo Librandi, May 22 2011

LinearRecurrence[{5,-10,10,-5,1},{1,120,947,3652,9985},30] (* Harvey P. Dale, May 04 2024 *)

a(n) = n*((145*n^3)-(280*n^2)+(179*n)-38)/6
a(n) = C(n+3,4) + 115 C(n+2,4) + 357 C(n+1,4) + 107 C(n,4)
a(n) = +5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5). G.f.: x*(1+115*x+357*x^2+107*x^3)/(1-x)^5. [R. J. Mathar, Jun 21 2010]

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A177342 a(n) = (4n^3-3n^2+5*n-3)/3.

Original entry on oeis.org

1, 9, 31, 75, 149, 261, 419, 631, 905, 1249, 1671, 2179, 2781, 3485, 4299, 5231, 6289, 7481, 8815, 10299, 11941, 13749, 15731, 17895, 20249, 22801, 25559, 28531, 31725, 35149, 38811, 42719, 46881, 51305, 55999, 60971, 66229, 71781, 77635

Offset: 1

Bruno Berselli, May 06 2010 - Nov 27 2010

This sequence is related to the fourth powers (A000583) by n^4 = n*a(n) - Sum_{i=1..n-1} a(i) - (n-1), with n>1.

Also, n*a(n) - Sum_{i=1..n-1} a(i) provides the first column of A162624 and the second column of A162622 (or A162623). - Bruno Berselli, revised Dec 14 2012

B. Berselli, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

First differences: 2*A084849.
Partial sums: A178073.
Cf. A174723, A162622-A162624.

[(4*n^3-3*n^2+5*n-3)/3: n in [1..39]]; // Bruno Berselli, Aug 24 2011

I:=[1,9,31,75]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 19 2013

CoefficientList[Series[(1 + 5 x + x^2 + x^3) / (1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
Table[(4 n^3 - 3 n^2 + 5 n - 3)/3, {n, 1, 40}] (* Bruno Berselli, Feb 17 2015 *)
LinearRecurrence[{4,-6,4,-1},{1,9,31,75},40] (* Harvey P. Dale, Jul 31 2021 *)

a(n)=(4*n^3-3*n^2+5*n-3)/3 \\ Charles R Greathouse IV, Jun 23 2011

G.f.: x*(1 + 5*x + x^2 + x^3)/(1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) - a(-n) = 2*A004006(2n).
a(n) + a(-n) = -A002522(n).
a(n) = 1 + (n-1)*(4*n^2+n+6)/3 = 2*A174723(n)-1.

Formulae added and revised by Bruno Berselli, Feb 17 2015

A266212 Positive integers x such that x^3 = y^4 + z^2 for some positive integers y and z.

Original entry on oeis.org

8, 13, 20, 25, 40, 125, 128, 193, 200, 208, 225, 313, 320, 328, 400, 500, 605, 640, 648, 1000, 1053, 1156, 1521, 1620, 1625, 1681, 1700, 2000, 2025, 2048, 2125, 2465, 2493, 2873, 2920, 3025, 3088, 3185, 3200, 3240, 3328, 3400, 3600, 3656, 3748, 3816, 4225, 4625, 4913, 5000, 5008, 5120, 5248, 6400, 6728, 6760, 6793, 6845, 7225, 8000

Offset: 1

Zhi-Wei Sun, Dec 23 2015

nonn

If x^3 = y^4 + z^2, then (a^(4k)*x)^3 = (a^(3k)*y)^4 + (a^(6k)*z)^2 for all a = 1,2,3,... and k = 0,1,2,... So the sequence has infinitely many terms.
Conjecture: For any integer m, there are infinitely many triples (x,y,z) of positive integers with x^4 - y^3 + z^2 = m.
This is stronger than the conjecture in A266152.

			a(1) = 8 since 8^3 = 4^4 + 16^2.
a(2) = 13 since 13^3 = 3^4 + 46^2.
a(3) = 20 since 20^3 = 4^4 + 88^2.
a(8) = 193 since 193^3 = 6^4 + 2681^2.
a(12) = 313 since 313^3 = 66^4 + 3419^2.
a(20) = 1000 since 1000^3 = 100^4 + 30000^2.

Zhi-Wei Sun and Chai Wah Wu, Table of n, a(n) for n = 1..698 n = 1..100 from Zhi-Wei Sun
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, A266152, A266153.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
n=0;Do[Do[If[SQ[x^3-y^4],n=n+1;Print[n," ",x];Goto[aa]],{y,1,x^(3/4)}];Label[aa];Continue,{x,1,8000}]

cp's OEIS Frontend

A016948 a(n) = (6*n + 3)^4.

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A016960 a(n) = (6*n + 4)^4.

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A016972 a(n) = (6*n + 5)^4.

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A024196 a(n) = 2nd elementary symmetric function of the first n+1 odd positive integers.

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A062075 a(n) = n^4 * 4^n.

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A092181 Figurate numbers based on the 24-cell (4-D polytope with Schlaefli symbol {3,4,3}).

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A092182 Figurate numbers based on the 600-cell (4-D polytope with Schlaefli symbol {3,3,5}).

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Magma

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A177342 a(n) = (4n^3-3n^2+5*n-3)/3.