cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A262955 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a pentagonal number (A000326) or twice a pentagonal number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 05 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 10, 35, 75, 134, 415, 515, 1465, 2365, 3515, 4140.

Examples

			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(10) = 1 since 10 = 0^4 + 4*5/2 + p_5(0).
a(35) = 1 since 35 = 1^4 + 4*5/2 + 2*p_5(3).
a(75) = 1 since 75 = 2^4 + 5*6/2 + 2*p_5(4).
a(134) = 1 since 134 = 2^4 + 1*2/2 + p_5(9).
a(415) = 1 since 415 = 0^4 + 21*22/2 + 2*p_5(8).
a(515) = 1 since 515 = 0^4 + 6*7/2 + 2*p_5(13).
a(1465) = 1 since 1465 = 5^4 + 35*36/2 + p_5(12).
a(2365) = 1 since 2365 = 5^4 + 8*9/2 + 2*p_5(24).
a(3515) = 1 since 3515 = 5^4 + 51*52/2 + 2*p_5(23).
a(4140) = 1 since 4140 = 1^4 + 90*91/2 + 2*p_5(4).
		

Crossrefs

Programs

  • Mathematica
    PenQ[n_]:=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0)
    PQ[n_]:=PenQ[n]||PenQ[n/2]
    Do[r=0;Do[If[PQ[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}]

A266153 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

3, 3, 2, 6, 13, 2, 3, 5, 5, 3, 28, 4, 15, 4, 10, 33, 3, 7, 5, 238, 31, 3, 4, 5, 3, 11, 4, 5, 21, 11, 6, 4, 17, 11, 5, 98, 7, 4, 4, 5, 147, 19, 5, 4, 5, 6, 4, 29, 75, 1011, 7, 9, 7, 4, 8, 6, 59, 47, 4, 5, 71, 4, 17, 45, 13, 7, 18, 9, 175, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 22 2015

Keywords

Comments

The conjecture in A266152 implies that a(n) > 0 for all n > 0.
It seems that a(n) < n*(n+4)/2 for all n > 1.

Examples

			a(1) = 3 since -1 = 1^4 - 3^3 + 5^2.
a(2) = 3 since -2 = 2^4 - 3^3 + 3^2.
a(11) = 28 since -11 = 5^4 - 28^3 + 146^2.
a(20) = 238 since -20 = 32^4 - 238^3 + 3526^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
    Do[y=Floor[n^(1/3)]+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,1,70}]

A038992 Number of sublattices of index n in generic 5-dimensional lattice.

Original entry on oeis.org

1, 31, 121, 651, 781, 3751, 2801, 11811, 11011, 24211, 16105, 78771, 30941, 86831, 94501, 200787, 88741, 341341, 137561, 508431, 338921, 499255, 292561, 1429131, 508431, 959171, 925771, 1823451, 732541, 2929531, 954305, 3309747, 1948705, 2750971, 2187581, 7168161, 1926221
Offset: 1

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Author

Keywords

References

  • Michael Baake, "Solution of the coincidence problem in dimensions d <= 4", in R. V. Moody, ed., Math. of Long-Range Aperiodic Order, Kluwer 1997, pp. 9-44.

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, #*DivisorSum[#, #*DivisorSum[#, #*DivisorSum[#, # &] &] &] &]; Array[a, 50] (* Jean-François Alcover, Dec 02 2015, after Joerg Arndt *)
    f[p_, e_] := Product[(p^(e + k) - 1)/(p^k - 1), {k, 1, 4}]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
  • PARI
    a(n)=sumdiv(n,x, x * sumdiv(x,y, y * sumdiv(y,z, z * sumdiv(z,w, w ) ) ) ); /* Joerg Arndt, Oct 07 2012 */

Formula

f(Q, n) = Sum_{d|n} d*f(Q-1, d); here Q=5.
Multiplicative with a(p^e) = Product_{k=1..4} (p^(e+k)-1)/(p^k-1).
Dirichlet g.f.: zeta(s)*zeta(s-1)*zeta(s-2)*zeta(s-3)*zeta(s-4). Dirichlet convolution of A038991 with A000583. - R. J. Mathar, Mar 31 2011
Sum_{k=1..n} a(k) ~ c * n^5, where c = Pi^6*zeta(3)*zeta(5)/2700 = 0.443822... . - Amiram Eldar, Oct 19 2022

Extensions

Offset changed from 0 to 1 by R. J. Mathar, Mar 31 2011
More terms from Joerg Arndt, Oct 07 2012

A219086 a(n) = floor((n + 1/2)^4).

Original entry on oeis.org

0, 5, 39, 150, 410, 915, 1785, 3164, 5220, 8145, 12155, 17490, 24414, 33215, 44205, 57720, 74120, 93789, 117135, 144590, 176610, 213675, 256289, 304980, 360300, 422825, 493155, 571914, 659750, 757335, 865365, 984560, 1115664
Offset: 0

Views

Author

Clark Kimberling, Jan 01 2013

Keywords

Comments

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/4 and { } = fractional part. Equivalently, the jump sequence of f(x) = x^(1/4), in the sense that these are the nonnegative integers k for which round(k^p) < round((k+1)^p). For details and a guide to related sequences, see A219085.
-4*a(n) gives the real part of (n+n*i)*((n+1)+n*i)*(n+(n+1)*i)*((n+1)+(n+1)*i). The imaginary part is always zero. - Jon Perry, Feb 05 2014
Numbers k such that 16*k+1 is a fourth power. - Bruno Berselli, May 29 2018
The row sums of "Floyd's Triangle", which is a triangular array of natural numbers beginning with the number 1, produce the sequence A006003. A006003 can be bisected to get the Rhombic Dodecahedron Sequence A005917, whose n-th partial sum is n^4, and A317297, whose n-th partial sum is a(n). Interleave n^4 or A000583 back with {a(n)} to get A011863, whose first differences are A019298. Finally, A011863(n)-A011863(n-2) = A006003(n-1). - Bruce J. Nicholson, Dec 22 2019

Examples

			0^(1/4) = 0.000...; 1^(1/4) = 1.000...
5^(1/4) = 1.495...; 6^(1/4) = 1.565...
39^(1/4) = 2.499...; 40^(1/4) = 2.514...
		

Crossrefs

Programs

Formula

G.f.: (5*x^3 + 14*x^2 + 5*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = (2*n^4 + 4*n^3 + 3*n^2 + n)/2. - J. M. Bergot, Apr 05 2014
a(n) = Sum_{i=0..n} i*(4*i^2 + 1) = n*(n + 1)*(2*n^2 + 2*n + 1)/2. - Bruno Berselli, Feb 09 2017
a(n) = lcm((2*n + 1)^2 - 1, (2*n + 1)^2 + 1)/8 for n>=1. - Lechoslaw Ratajczak, Mar 26 2017
a(n) = A000217(n) * A001844(n). - Bruce J. Nicholson, May 14 2017
E.g.f.: (1/2)*exp(x)*x*(10 + 29*x + 16*x^2 + 2*x^3). - Stefano Spezia, Dec 27 2019
a(n) = ((2*n+1)^4 - 1)/16. - Jianing Song, Jan 03 2023
Sum_{n>=1} 1/a(n) = 6 - 2*Pi*tanh(Pi/2). - Amiram Eldar, Jan 08 2023

A004831 Numbers that are the sum of at most 2 nonzero 4th powers.

Original entry on oeis.org

0, 1, 2, 16, 17, 32, 81, 82, 97, 162, 256, 257, 272, 337, 512, 625, 626, 641, 706, 881, 1250, 1296, 1297, 1312, 1377, 1552, 1921, 2401, 2402, 2417, 2482, 2592, 2657, 3026, 3697, 4096, 4097, 4112, 4177, 4352, 4721, 4802, 5392, 6497, 6561, 6562, 6577, 6642
Offset: 1

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Author

Keywords

Comments

Apart from 0, 1, 2, there are no three consecutive terms up to 10^16. The first two consecutive terms not of the form n^4, n^4+1 are 3502321 = 25^4 + 42^4, 3502322 = 17^4 + 43^4. - Charles R Greathouse IV, Oct 17 2017

Crossrefs

Subsequences include A003336, A000583 and A002645.

Programs

  • Haskell
    a004831 n = a004831_list !! (n-1)
    a004831_list = [x ^ 4 + y ^ 4 | x <- [0..], y <- [0..x]]
    -- Reinhard Zumkeller, Jul 15 2013
    
  • Mathematica
    Reap[For[n = 0, n < 10000, n++, If[MatchQ[ PowersRepresentations[n, 2, 4], {{, },_}], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 30 2017 *)
  • PARI
    is(n)=#thue(thueinit(z^4+1),n) \\ Ralf Stephan, Oct 18 2013
    
  • PARI
    list(lim)=my(v=List(),t); for(m=0,sqrtnint(lim\=1,4), for(n=0, min(sqrtnint(lim-m^4,4),m), listput(v,n^4+m^4))); Set(v) \\ Charles R Greathouse IV, Sep 28 2015

Formula

Call f(x) the number of terms if this sequence up to x. Then x^(7/16) << f(x) << x^(1/2); in other words, n^2 << a(n) << n^(16/7). The upper bound becomes O(n^2) if A230562 is finite. - Charles R Greathouse IV, Jul 12 2024

A007204 Crystal ball sequence for D_4 lattice.

Original entry on oeis.org

1, 25, 169, 625, 1681, 3721, 7225, 12769, 21025, 32761, 48841, 70225, 97969, 133225, 177241, 231361, 297025, 375769, 469225, 579121, 707281, 855625, 1026169, 1221025, 1442401, 1692601, 1974025, 2289169, 2640625, 3031081, 3463321, 3940225, 4464769, 5040025
Offset: 0

Views

Author

N. J. A. Sloane and J. H. Conway, Apr 28 1994

Keywords

Comments

Equals binomial transform of [1, 24, 120, 192, 96, 0, 0, 0, ...]. - Gary W. Adamson, Aug 13 2009
Hypotenuse of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + A069074(n-1)^2 = a(n)^2. - Martin Renner, Nov 12 2011
Numbers n such that n*x^4 + x^2 + 1 is reducible. - Arkadiusz Wesolowski, Nov 04 2013

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 53.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [(2*n^2+2*n+1)^2: n in [0..40]]; // Vincenzo Librandi, Nov 18 2016
    
  • Maple
    A007204:=n->(2*n^2+2*n+1)^2; seq(A007204(n), n=0..30);
  • Mathematica
    Table[(2n^2+2n+1)^2,{n,0,30}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,25,169,625,1681},40] (* Harvey P. Dale, Mar 03 2013 *)
  • PARI
    a(n)=(2*n^2+2*n+1)^2 \\ Charles R Greathouse IV, Feb 08 2017

Formula

G.f.: (1 + 54*x^2 + 20*x + 20*x^3 + x^4)/(1-x)^5.
a(0)=1, a(1)=25, a(2)=169, a(3)=625, a(4)=1681, a(n)=5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Mar 03 2013
Sum_{n>=0} 1/a(n) = Pi*(sinh(Pi) - Pi)/(2*(cosh(Pi) + 1)) = 1.0487582722070177... - Ilya Gutkovskiy, Nov 18 2016
a(n) = A016754(n) + A060300(n). - Bruce J. Nicholson, Apr 14 2017
a(n) = A001844(n)^2 = (2*n^2+2*n+1)^2. - Bruce J. Nicholson, May 15 2017
a(n) = A000583(n+1) + A099761(n) + 2*A005563(n-1)*A000290(n). - Charlie Marion, Jan 14 2021
E.g.f.: exp(x)*(1 + 24*x + 60*x^2 + 32*x^3 + 4*x^4). - Stefano Spezia, Jun 06 2021

Extensions

More terms from Harvey P. Dale, Mar 03 2013

A016780 a(n) = (3*n+1)^4.

Original entry on oeis.org

1, 256, 2401, 10000, 28561, 65536, 130321, 234256, 390625, 614656, 923521, 1336336, 1874161, 2560000, 3418801, 4477456, 5764801, 7311616, 9150625, 11316496, 13845841, 16777216, 20151121, 24010000, 28398241, 33362176, 38950081, 45212176, 52200625, 59969536, 68574961
Offset: 0

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Author

Keywords

Crossrefs

Cf. A000583 (n^4), A016777 (3n+1), A016778, A016779, A016781.

Programs

  • Magma
    [(3*n+1)^4: n in [0..30]]; // Vincenzo Librandi, Sep 21 2011
  • Mathematica
    (3*Range[0,30]+1)^4 (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,256,2401,10000,28561},30] (* Harvey P. Dale, Oct 21 2015 *)

Formula

From Harvey P. Dale, Oct 21 2015: (Start)
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5).
G.f.: -((1+251*x+1131*x^2+545*x^3+16*x^4)/(-1+x)^5). (End)
a(n) = A000583(A016777(n)). - Michel Marcus, Nov 06 2015
E.g.f.: exp(x)*(1+255*x+945*x^2+594*x^3+81*x^4). - Wolfdieter Lang, Apr 02 2017
Sum_{n>=0} 1/a(n) = PolyGamma(3, 1/3)/486. - Amiram Eldar, Mar 29 2022

A085537 a(n) = n^4 - n^3.

Original entry on oeis.org

0, 0, 8, 54, 192, 500, 1080, 2058, 3584, 5832, 9000, 13310, 19008, 26364, 35672, 47250, 61440, 78608, 99144, 123462, 152000, 185220, 223608, 267674, 317952, 375000, 439400, 511758, 592704, 682892, 783000, 893730, 1015808, 1149984, 1297032, 1457750, 1632960
Offset: 0

Views

Author

N. J. A. Sloane, Jul 05 2003

Keywords

Comments

For n>=1, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n} such that for a fixed x in {1,2,3,4} and a fixed y in {1,2,...,n} we have f(x)<>y. - Aleksandar M. Janjic and Milan Janjic, Mar 13 2007
Let K_n denote the complete graph on n (n>1) vertices. The sequence corresponds to the Wiener index of K_n X K_n (Cartesian product of K_n with itself). - K.V.Iyer, Mar 12 2009
Lewis proved that the order of a solvable nonabelian finite group |G| is less than or equal to e^4 - e^3, where when d is an irreducible character degree of G, then there is a positive integer e such that |G| = d(d+e). - Jonathan Vos Post, Jun 21 2012

Crossrefs

A diagonal of A228273.
Cf. A085540 (same sequence with initial 0 dropped).

Programs

  • Mathematica
    Table[(n - 1) n^3, {n, 0, 20}] (* Eric W. Weisstein, Sep 08 2017 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 54, 192, 500}, {0, 20}] (* Eric W. Weisstein, Sep 08 2017 *)
    CoefficientList[Series[2 x^2 (4 + 7 x + x^2)/(1 - x)^5, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *)
  • PARI
    A085537(n) = n^4-n^3

Formula

From R. J. Mathar, Sep 12 2008: (Start)
a(n) = A085540(n-1).
G.f.: 2*x^2*(4 + 7*x + x^2)/(1-x)^5. (End)
a(n) = A000583(n) - A000578(n). - Omar E. Pol, Jun 23 2012
Sum_{n>=2} 1/a(n) = 3 - zeta(2) - zeta(3) = A152419. - Daniel Suteu, Feb 06 2017
a(n) = 2*A092364(n+1). - Bruno Berselli, Sep 08 2017
Sum_{n>=2} (-1)^n/a(n) = Pi^2/12 + 2*log(2) + 3*zeta(3)/4 - 3. - Amiram Eldar, Jul 05 2020
E.g.f.: exp(x)*x^2*(4 + 5*x + x^2). - Stefano Spezia, Jul 06 2021
Product_{n>=2} (1 - 1/a(n)) = A146489. - Amiram Eldar, Nov 22 2022

A100019 a(n) = n^4 + n^3 + n^2.

Original entry on oeis.org

0, 3, 28, 117, 336, 775, 1548, 2793, 4672, 7371, 11100, 16093, 22608, 30927, 41356, 54225, 69888, 88723, 111132, 137541, 168400, 204183, 245388, 292537, 346176, 406875, 475228, 551853, 637392, 732511, 837900, 954273, 1082368, 1222947
Offset: 0

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Nov 19 2004

Keywords

Comments

a(n) are the numbers m such that: j^2 = j + m + sqrt(j*m) with corresponding numbers j given by A002061(n+1), and with sqrt(j*m) = A027444(n) = n* A002061(n+1). - Richard R. Forberg, Sep 03 2013.

Crossrefs

Programs

Formula

From Indranil Ghosh, Apr 15 2017: (Start)
G.f.: -x(3 + 13x + 7x^2 + x^3)/(x - 1)^5
E.g.f.: exp(x)*x*(3 + 11x + 7x^2 + x^3)
(End)

A141046 a(n) = 4*n^4.

Original entry on oeis.org

0, 4, 64, 324, 1024, 2500, 5184, 9604, 16384, 26244, 40000, 58564, 82944, 114244, 153664, 202500, 262144, 334084, 419904, 521284, 640000, 777924, 937024, 1119364, 1327104, 1562500, 1827904, 2125764, 2458624, 2829124, 3240000, 3694084, 4194304, 4743684, 5345344
Offset: 0

Views

Author

Fredrik Johansson, Jul 31 2008

Keywords

Comments

Nonnegative integers a(n) such that (-a(n))^(1/4) is a Gaussian integer, since (n + n*i)^4 = -4*n^4
For n > 1, a(n) + k^4 is not prime for any k. - Derek Orr, May 31 2014
Suppose the vertices of a triangle are (T(n), T(n+j)), (T(n+2*j), T(n+3*j)) and (T(n+4*j), T(n+5*j)) where T(n) is the n-th triangular number. Then the area of this triangle will be a(j). - Charlie Marion, Mar 06 2021

Crossrefs

Programs

Formula

a(n) = 4*n^4.
a(n) = A008586(A000583(n)) = A000290(A005843(A000290(n))). - Reinhard Zumkeller, Jan 25 2012
G.f.: 4*x*(1 + x)*(1 + 10*x + x^2)/(1 - x)^5. - Chai Wah Wu, Jun 22 2016
From G. C. Greubel, Jun 22 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
E.g.f.: 4*x*(1 + 7*x + 6*x^2 + x^3)*exp(x). (End)
a(n) = A001105(n)^2. - Bruce J. Nicholson, Apr 03 2017
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^4/360.
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/2880.
Product_{n>=1} (1 + 1/a(n)) = 2*cosh(Pi/2)^2/Pi^2.
Product_{n>=1} (1 - 1/a(n)) = 2*sin(Pi/sqrt(2))*sinh(Pi/sqrt(2))/Pi^2. (End)
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