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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A028916 Friedlander-Iwaniec primes: Primes of form a^2 + b^4.

Original entry on oeis.org

2, 5, 17, 37, 41, 97, 101, 137, 181, 197, 241, 257, 277, 281, 337, 401, 457, 577, 617, 641, 661, 677, 757, 769, 821, 857, 881, 977, 1097, 1109, 1201, 1217, 1237, 1297, 1301, 1321, 1409, 1481, 1601, 1657, 1697, 1777, 2017, 2069, 2137, 2281, 2389, 2417, 2437
Offset: 1

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Author

Keywords

Comments

John Friedlander and Henryk Iwaniec proved that there are infinitely many such primes.
A256852(A049084(a(n))) > 0. - Reinhard Zumkeller, Apr 11 2015
Primes in A111925. - Robert Israel, Oct 02 2015
Its intersection with A185086 is A262340, by the uniqueness part of Fermat's two-squares theorem. - Jonathan Sondow, Oct 05 2015
Cunningham calls these semi-quartan primes. - Charles R Greathouse IV, Aug 21 2017
Primes of the form (x^2 + y^2)/2, where x > y > 0, such that (x-y)/2 or (x+y)/2 is square. - Thomas Ordowski, Dec 04 2017
Named after the Canadian mathematician John Benjamin Friedlander (b. 1941) and the Polish-American mathematician Henryk Iwaniec (b. 1947). - Amiram Eldar, Jun 19 2021

Examples

			2 = 1^2 + 1^4.
5 = 2^2 + 1^4.
17 = 4^2 + 1^4 = 1^2 + 2^4.
		

Crossrefs

Cf. A000290, A000583, A000040, A256852, A256863 (complement), A002645 (subsequence), subsequence of A247857.
Primes of form n^2 + b^4, b fixed: A002496 (b = 1), A243451 (b = 2), A256775 (b = 3), A256776 (b = 4), A256777 (b = 5), A256834 (b = 6), A256835 (b = 7), A256836 (b = 8), A256837 (b = 9), A256838 (b = 10), A256839 (b = 11), A256840 (b = 12), A256841 (b = 13).

Programs

  • Haskell
    a028916 n = a028916_list !! (n-1)
    a028916_list = map a000040 $ filter ((> 0) . a256852) [1..]
    -- Reinhard Zumkeller, Apr 11 2015
  • Maple
    N:= 10^5: # to get all terms <= N
    S:= {seq(seq(a^2+b^4, a = 1 .. floor((N-b^4)^(1/2))),b=1..floor(N^(1/4)))}:
    sort(convert(select(isprime,S),list)); # Robert Israel, Oct 02 2015
  • Mathematica
    nn = 10000; t = {}; Do[n = a^2 + b^4; If[n <= nn && PrimeQ[n], AppendTo[t, n]], {a, Sqrt[nn]}, {b, nn^(1/4)}]; Union[t] (* T. D. Noe, Aug 06 2012 *)
  • PARI
    list(lim)=my(v=List([2]),t);for(a=1,sqrt(lim\=1),forstep(b=a%2+1, sqrtint(sqrtint(lim-a^2)), 2, t=a^2+b^4;if(isprime(t),listput(v,t)))); vecsort(Vec(v),,8) \\ Charles R Greathouse IV, Jun 12 2013
    

Extensions

Title expanded by Jonathan Sondow, Oct 02 2015

A003345 Numbers that are the sum of 11 positive 4th powers.

Original entry on oeis.org

11, 26, 41, 56, 71, 86, 91, 101, 106, 116, 121, 131, 136, 146, 151, 161, 166, 171, 176, 181, 186, 196, 201, 211, 216, 226, 231, 241, 246, 251, 261, 266, 276, 281, 291, 296, 306, 311, 326, 331, 341, 346, 356, 361, 371, 376, 386, 391, 401, 406, 411, 416, 421, 426, 436
Offset: 1

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Author

Keywords

Examples

			From _David A. Corneth_, Aug 03 2020: (Start)
4422 is in the sequence as 4422 = 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.
6611 is in the sequence as 6611 = 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4.
7286 is in the sequence as 7286 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 9^4. (End)
		

Crossrefs

Cf. A000583 (4th powers).

Programs

  • Mathematica
    With[{nn=5},Select[Union[Total/@Tuples[Range[nn]^4,11]],#<=nn^4-10&]] (* Harvey P. Dale, Jul 20 2022 *)

Extensions

Incorrect program removed by David A. Corneth, Aug 03 2020

A014820 a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.

Original entry on oeis.org

1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481
Offset: 0

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Author

Keywords

Comments

a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009
If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008

References

  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Crossrefs

Programs

  • GAP
    List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019
  • Magma
    [(1/3)*(n^2+2*n+3)*(n+1)^2: n in [0..40]]; // Vincenzo Librandi, May 22 2011
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n),r=0..d-1); end; [seq(be(4,n),n=0..100)];
  • Mathematica
    LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)
  • PARI
    a(n)=(n+1)^2*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Apr 17 2012
    
  • R
    a <- c(1, 8, 33, 96,225)
    for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
    a # Yosu Yurramendi, Sep 03 2013
    
  • Sage
    [((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019
    

Formula

Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009
From Vladeta Jovovic, Apr 03 2002: (Start)
G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013
From Bruce J. Nicholson, Jan 23 2019: (Start)
Sum_{i=0..n} a(i) = A061927(n+1).
a(n) = 4*A002415(n+1) + A000290(n+1) = A039623(n+1) + A002415(n+1). (End)
E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024

Extensions

Formula index corrected by R. J. Mathar, Jul 18 2009

A262827 Number of ordered ways to write n as w^2 + x^3 + y^4 + 2*z^4, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

1, 3, 4, 4, 4, 3, 2, 2, 2, 3, 4, 4, 4, 2, 1, 1, 2, 5, 5, 5, 4, 1, 1, 1, 2, 4, 5, 6, 6, 3, 3, 2, 3, 7, 6, 4, 4, 5, 4, 3, 3, 4, 5, 4, 5, 4, 3, 2, 2, 8, 5, 3, 6, 4, 3, 2, 2, 5, 4, 4, 5, 2, 1, 2, 5, 9, 7, 5, 7, 4, 3, 1, 2, 4, 3, 5, 5, 2, 1, 3, 3, 8, 9, 8, 8, 5, 2, 1, 2, 5, 6, 7, 7, 3, 2, 2, 4, 7, 7, 2, 7
Offset: 0

Views

Author

Zhi-Wei Sun, Oct 03 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0. Also, a(n) = 1 only for the following 41 values of n: 0, 14, 15, 21, 22, 23, 62, 71, 78, 87, 136, 216, 405, 437, 448, 477, 535, 583, 591, 623, 671, 696, 885, 950, 1046, 1135, 1206, 1208, 1248, 1317, 2288, 2383, 2543, 3167, 3717, 3974, 6847, 7918, 8328, 9096, 21935.
We have verified that a(n) > 0 for all n = 0..10^7.
Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 07 2023
We also conjecture that if f(w,x,y,z) is one of the 8 polynomials 2w^2+x^3+4y^3+z^4, w^2+x^3+2y^3+c*z^3 (c = 3,4,5,6) and w^2+x^3+2y^3+d*z^4 (d = 1,3,6) then each n = 0,1,2,... can be written as f(w,x,y,z) with w,x,y,z nonnegative integers. - Zhi-Wei Sun, Dec 30 2017
Conjecture verified up to 10^11 for all 8 polynomials. - Mauro Fiorentini, Jul 07 2023

Examples

			a(14) = 1 since 14 = 2^2 + 2^3 + 0^4 + 2*1^4.
a(87) = 1 since 87 = 2^2 + 0^3 + 3^4 + 2*1^4.
a(216) = 1 since 216 = 0^2 + 6^3 + 0^4 + 2*0^4.
a(405) = 1 since 405 = 18^2 + 0^3 + 3^4 + 2*0^4.
a(1248) = 1 since 1248 = 31^2 + 5^3 + 0^4 + 2*3^4.
a(1317) = 1 since 1317 = 23^2 + 1^3 + 5^4 + 2*3^4.
a(2288) = 1 since 2288 = 44^2 + 4^3 + 4^4 +2*2^4.
a(2383) = 1 since 2383 = 1462 + 9^3 + 6^4 + 2*3^4.
a(2543) = 1 since 2543 = 50^2 + 3^3 + 2^4 + 2*0^4.
a(3167) = 1 since 3167 = 54^2 + 2^3 + 3^4 + 2*3^4.
a(3717) = 1 since 3717 = 18^2 + 15^3 + 2^4 + 2*1^4.
a(3974) = 1 since 3974 = 39^2 + 13^3 + 4^4 + 2*0^4.
a(6847) = 1 since 6847 = 52^2 + 15^3 + 4^4 + 2*4^4.
a(7918) = 1 since 7918 = 46^2 + 10^3 + 0^4 + 2*7^4.
a(8328) = 1 since 8328 = 42^2 + 1^3 + 9^4 + 2*1^4.
a(9096) = 1 since 9096 = 44^2 + 18^3 + 6^4 + 2*2^4.
a(21935) = 1 since 21935 = 66^2 + 26^3 + 1^4 + 2*1^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^3-y^4-2*z^4],r=r+1],{x,0,n^(1/3)},{y,0,(n-x^3)^(1/4)},{z,0,((n-x^3-y^4)/2)^(1/4)}];Print[n," ",r];Continue,{n,0,100}]

A270969 Number of ways to write n as w^4 + x^2 + y^2 + z^2, where w, x, y and z are nonnegative integers with x <= y <= z.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 1, 1, 3, 3, 2, 2, 2, 2, 1, 2, 4, 5, 4, 3, 3, 3, 1, 2, 5, 5, 5, 3, 3, 4, 1, 2, 5, 6, 4, 4, 4, 4, 2, 2, 6, 6, 4, 2, 5, 4, 1, 2, 5, 7, 6, 5, 4, 7, 3, 2, 6, 4, 4, 3, 4, 5, 2, 2, 6, 9, 6, 4, 6, 6, 1, 3, 6, 6, 7, 3, 5, 5, 1, 2
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 27 2016

Keywords

Comments

Theorem: a(n) > 0 for all n = 0,1,2,.... In other words, any nonnegative integer can be written as the sum of a fourth power and three squares.
This is stronger than Lagrange's four-square theorem, and it can be proved by induction on n. It is easy to check that a(n) > 0 for all n = 0..16. Now let n be an integer greater than 16, and assume that a(m) > 0 for all m = 0..n-1. If 16|n, then n/16 can be written as w^4+x^2+y^2+z^2 with w,x,y,z integers, and hence n = (2w)^4+(4x)^2+(4y)^2+(4z)^2. If n == 8 (mod 16), then n is not of the form 4^k*(8q+7) and hence n = 0^4+x^2+y^2+z^2 for some integers x,y,z. If n == 4 (mod 8), then n-1^4 can be written as the sum of three squares. If n == 2 (mod 4), then n-0^4 is a sum of three squares. If n == 7 (mod 8), then n-1^4 can be written as the sum of three squares. If n is odd but not congruent to 7 modulo 8, then n-0^4 can be expressed as the sum of three squares.
We have a(n) = 1 if n has the form 16^k*q with k a nonnegative integer and q among 7, 8, 15, 23, 31, 47, 71, 79. In fact, if n = 16*m with m > 0, and 16*m = w^4+x^2+y^2+z^2 with w,x,y,z integers, then w,x,y,z are all even and hence m = (w/2)^4+(x/2)^2+(y/2)^2+(z/2)^2. Therefore a(16*m) = a(m) for all m > 0. It is easy to check that a(q) = 1 for every q = 7, 8, 15, 23, 31, 47, 71, 79.
For (a,b,c) = (1,1,2),(1,1,3),(1,1,4),(1,1,6),(1,2,2),(1,2,3),(1,2,4),(1,2,5), we are also able to show that any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.
Conjecture: For each triple (a,b,c) = (1,2,11),(1,2,12),(1,2,13),(2,3,5), any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.

Examples

			a(7) = 1 since 7 = 1^4 + 1^2 + 1^2 + 2^2.
a(8) = 1 since 8 = 0^4 + 0^2 + 2^2 + 2^2.
a(15) = 1 since 15 = 1^4 + 1^2 + 2^2 + 3^2.
a(23) = 1 since 23 = 1^4 + 2^2 + 3^2 + 3^2.
a(31) = 1 since 31 = 1^4 + 1^2 + 2^2 + 5^2.
a(47) = 1 since 47 = 1^4 + 1^2 + 3^2 + 6^2.
a(71) = 1 since 71 = 1^4 + 3^2 + 5^2 + 6^2.
a(79) = 1 since 79 = 1^4 + 2^2 + 5^2 + 7^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-w^4-x^2-y^2],r=r+1],{w,0,n^(1/4)},{x,0,Sqrt[(n-w^4)/3]},{y,x,Sqrt[(n-w^4-x^2)/2]}];Print[n," ",r];Continue,{n,0,80}]

A357641 Number of integer compositions of 2n whose half-alternating sum is 0.

Original entry on oeis.org

1, 0, 2, 8, 28, 104, 396, 1504, 5720, 21872, 83980, 323344, 1248072, 4828784, 18721080, 72711552, 282861360, 1101980000, 4298748300, 16789002736, 65641204200, 256895795312, 1006308200040, 3945185586368, 15478849767888, 60774329914144, 238775589937976
Offset: 0

Views

Author

Gus Wiseman, Oct 12 2022

Keywords

Comments

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 - ...

Examples

			The a(0) = 1 through a(3) = 8 compositions:
  ()  .  (112)   (123)
         (1111)  (213)
                 (1212)
                 (1221)
                 (2112)
                 (2121)
                 (11121)
                 (11211)
		

Crossrefs

The skew-alternating version appears to be A001700.
The version for partitions is A035363.
The skew-alternating form is A088218 (also for full alternating sum).
These compositions are ranked by A357625, reverse A357626.
For reversed partitions we have A357639, ranked by A357631.
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.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<3, [1, 0, 2][n+1],
          (8*(n-3)*(5*n-7)*(2*n-5)*a(n-3) -4*(5*n-12)*(n-2)^2*a(n-2)
           +2*(2*n-5)*(5*n-7)*n*a(n-1))/((5*n-12)*(n+1)*(n-2)))
        end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Oct 19 2022
  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[2n],halfats[#]==0&]],{n,0,7}]

Extensions

a(11)-a(26) from Alois P. Heinz, Oct 19 2022

A008456 12th powers: a(n) = n^12.

Original entry on oeis.org

0, 1, 4096, 531441, 16777216, 244140625, 2176782336, 13841287201, 68719476736, 282429536481, 1000000000000, 3138428376721, 8916100448256, 23298085122481, 56693912375296, 129746337890625, 281474976710656, 582622237229761
Offset: 0

Views

Author

Keywords

Comments

Numbers which are square, cubic and quartic. - Doug Bell, Jun 03 2017

Crossrefs

a(n) = A123868(n) + 1.
Cf. A000290 (squares), A000578 (cubes), A000583 (4th powers), A001014 (6th powers), A008454 (10th powers), A008455 (11th powers), A010801 (13th powers).
Cf. A013670 (zeta(12)).

Programs

Formula

Multiplicative with a(p^e) = p^(12*e). - David W. Wilson, Aug 01 2001
a(n) = A000290(n)^6 = A000578(n)^4 = A000583(n)^3 = A001014(n)^2. - Doug Bell, Jun 03 2017
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(12) = 691*Pi^12/638512875 (A013670).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2047*zeta(12)/2048 = 1414477*Pi^12/1307674368000. (End)
a(n) = 13*a(n-1)-78*a(n-2)+286*a(n-3)-715*a(n-4)+1287*a(n-5)-1716*a(n-6)+1716*a(n-7)-1287*a(n-8)+715*a(n-9)-286*a(n-10)+78*a(n-11)-13*a(n-12)+a(n-13). - Wesley Ivan Hurt, Dec 02 2021
Intersection of A000578 and A000583; i.e., cubes and 4th powers. - M. F. Hasler, Jul 03 2025

A272351 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x^4 + 8*y*z*(y^2+z^2) a fourth power, where w,x,y,z are nonnegative integers with w >= x and y > z.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 2, 3, 2, 1, 4, 5, 2, 2, 3, 3, 2, 2, 3, 5, 4, 1, 5, 6, 1, 1, 5, 4, 5, 3, 2, 5, 2, 3, 7, 7, 3, 2, 5, 4, 2, 1, 5, 8, 7, 2, 5, 9, 1, 3, 4, 4, 5, 2, 5, 8, 6, 1, 8, 8, 4, 4, 6, 5, 1, 5, 5, 10, 6, 2, 6, 8, 1, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 27 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k, 4^k*m (k = 0,1,2,... and m = 3, 7, 31, 55, 71, 79, 151, 191).
(ii) For (b,c) = (8,8), (16,64), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x^4 + b*y^3*z + c*y*z^3 a fourth power, where w is a positive integer and x,y,z are nonnegative integers.
(iii) For each triple (a,b,c) = (1,20,60), (1,24,56), (9,20,60), (9,32,96), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with a*x^4 + b*y^3*z + c*y*z^3 a square, where w is a positive integer and x,y,z are nonnegative integers.
The author has proved part (ii) of the conjecture in arXiv:1604.06723. - Zhi-Wei Sun, May 09 2016

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 = 0, 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4.
a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4.
a(3) = 1 since 3 = 1^2 + 1^2 + 1^2 + 0^2 with 1 = 1, 1 > 0 and 1^4 + 8*1*0*(1^2+0^2) = 1^4.
a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1 = 1, 3 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.
a(31) = 1 since 31 = 5^2 + 1^2 + 2^2 + 1^2 with 5 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.
a(55) = 1 since 55 = 7^2 + 1^2 + 2^2 + 1^2 with 7 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 3 > 1, 6 > 5 and 1^4 + 8*6*5*(6^2+5^2) = 11^4.
a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5 > 3, 6 > 3 and 3^4 + 8*6*3*(6^2+3^2) = 9^4.
a(151) = 1 since 151 = 5^2 + 3^2 + 9^2 + 6^2 with 5 > 3, 9 > 6 and 3^4 + 8*9*6*(9^2+6^2) = 15^4.
a(191) = 1 since 191 = 3^2 + 1^2 + 10^2 + 9^2 with 3 > 1, 10 > 9 and 1^4 + 8*10*9*(10^2+9^2) = 19^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&QQ[x^4+8y*z*(y^2+z^2)],r=r+1],{z,0,(Sqrt[2n-1]-1)/2},{y,z+1,Sqrt[n-z^2]},{x,0,Sqrt[(n-y^2-z^2)/2]}];Print[n," ",r];Continue,{n,1,80}]

A003992 Square array read by upwards antidiagonals: T(n,k) = n^k for n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 9, 8, 1, 0, 1, 5, 16, 27, 16, 1, 0, 1, 6, 25, 64, 81, 32, 1, 0, 1, 7, 36, 125, 256, 243, 64, 1, 0, 1, 8, 49, 216, 625, 1024, 729, 128, 1, 0, 1, 9, 64, 343, 1296, 3125, 4096, 2187, 256, 1, 0, 1, 10, 81, 512, 2401, 7776, 15625, 16384, 6561, 512, 1, 0
Offset: 0

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Author

Keywords

Comments

If the array is transposed, T(n,k) is the number of oriented rows of n colors using up to k different colors. The formula would be T(n,k) = [n==0] + [n>0]*k^n. The generating function for column k would be 1/(1-k*x). For T(3,2)=8, the rows are AAA, AAB, ABA, ABB, BAA, BAB, BBA, and BBB. - Robert A. Russell, Nov 08 2018
T(n,k) is the number of multichains of length n from {} to [k] in the Boolean lattice B_k. - Geoffrey Critzer, Apr 03 2020

Examples

			Rows begin:
[1, 0,  0,   0,    0,     0,      0,      0, ...],
[1, 1,  1,   1,    1,     1,      1,      1, ...],
[1, 2,  4,   8,   16,    32,     64,    128, ...],
[1, 3,  9,  27,   81,   243,    729,   2187, ...],
[1, 4, 16,  64,  256,  1024,   4096,  16384, ...],
[1, 5, 25, 125,  625,  3125,  15625,  78125, ...],
[1, 6, 36, 216, 1296,  7776,  46656, 279936, ...],
[1, 7, 49, 343, 2401, 16807, 117649, 823543, ...], ...
		

Crossrefs

Main diagonal is A000312. Other diagonals include A000169, A007778, A000272, A008788. Antidiagonal sums are in A026898.
Cf. A099555.
Transpose is A004248. See A051128, A095884, A009999 for other versions.
Cf. A277504 (unoriented), A293500 (chiral).

Programs

  • Magma
    [[(n-k)^k: k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 08 2018
  • Mathematica
    Table[If[k == 0, 1, (n - k)^k], {n, 0, 11}, {k, 0, n}]//Flatten
  • PARI
    T(n,k) = (n-k)^k \\ Charles R Greathouse IV, Feb 07 2017
    

Formula

E.g.f.: Sum T(n,k)*x^n*y^k/k! = 1/(1-x*exp(y)). - Paul D. Hanna, Oct 22 2004
E.g.f.: Sum T(n,k)*x^n/n!*y^k/k! = e^(x*e^y). - Franklin T. Adams-Watters, Jun 23 2006

Extensions

More terms from David W. Wilson
Edited by Paul D. Hanna, Oct 22 2004

A297845 Encoded multiplication table for polynomials in one indeterminate with nonnegative integer coefficients. Symmetric square array T(n, k) read by antidiagonals, n > 0 and k > 0. See comment for details.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 9, 9, 5, 1, 1, 6, 7, 16, 7, 6, 1, 1, 7, 15, 25, 25, 15, 7, 1, 1, 8, 11, 36, 11, 36, 11, 8, 1, 1, 9, 27, 49, 35, 35, 49, 27, 9, 1, 1, 10, 25, 64, 13, 90, 13, 64, 25, 10, 1, 1, 11, 21, 81, 125, 77, 77, 125, 81
Offset: 1

Views

Author

Rémy Sigrist, Jan 10 2018

Keywords

Comments

For any number n > 0, let f(n) be the polynomial in a single indeterminate x where the coefficient of x^e is the prime(1+e)-adic valuation of n (where prime(k) denotes the k-th prime); f establishes a bijection between the positive numbers and the polynomials in a single indeterminate x with nonnegative integer coefficients; let g be the inverse of f; T(n, k) = g(f(n) * f(k)).
This table has many similarities with A248601.
For any n > 0 and m > 0, f(n * m) = f(n) + f(m).
Also, f(1) = 0 and f(2) = 1.
The function f can be naturally extended to the set of positive rational numbers: if r = u/v (not necessarily in reduced form), then f(r) = f(u) - f(v); as such, f is a homomorphism from the multiplicative group of positive rational numbers to the additive group of polynomials of a single indeterminate x with integer coefficients.
See A297473 for the main diagonal of T.
As a binary operation, T(.,.) is related to A306697(.,.) and A329329(.,.). When their operands are terms of A050376 (sometimes called Fermi-Dirac primes) the three operations give the same result. However the rest of the multiplication table for T(.,.) can be derived from these results because T(.,.) distributes over integer multiplication (A003991), whereas for A306697 and A329329, the equivalent derivation uses distribution over A059896(.,.) and A059897(.,.) respectively. - Peter Munn, Mar 25 2020
From Peter Munn, Jun 16 2021: (Start)
The operation defined by this sequence can be extended to be the multiplicative operator of a ring over the positive rationals that is isomorphic to the polynomial ring Z[x]. The extended function f (described in the author's original comments) is the isomorphism we use, and it has the same relationship with the extended operation that exists between their unextended equivalents.
Denoting this extension of T(.,.) as t_Q(.,.), we get t_Q(n, 1/k) = t_Q(1/n, k) = 1/T(n, k) and t_Q(1/n, 1/k) = T(n, k) for positive integers n and k. The result for other rationals is derived from the distributive property: t_Q(q, r*s) = t_Q(q, r) * t_Q(q, s); t_Q(q*r, s) = t_Q(q, s) * t_Q(r, s). This may look unusual because standard multiplication of rational numbers takes on the role of the ring's additive group.
There are many OEIS sequences that can be shown to be a list of the integers in an ideal of this ring. See the cross-references.
There are some completely additive sequences that similarly define by extension completely additive functions on the positive rationals that can be shown to be homomorphisms from this ring onto the integer ring Z, and these functions relate to some of the ideals. For example, the extended function of A048675, denoted A048675_Q, maps i/j to A048675(i) - A048675(j) for positive integers i and j. For any positive integer k, the set {r rational > 0 : k divides A048675_Q(r)} forms an ideal of the ring; for k=2 and k=3 the integers in this ideal are listed in A003159 and A332820 respectively.
(End)

Examples

			Array T(n, k) begins:
  n\k|  1   2   3    4    5    6    7     8    9    10
  ---+------------------------------------------------
    1|  1   1   1    1    1    1    1     1    1     1  -> A000012
    2|  1   2   3    4    5    6    7     8    9    10  -> A000027
    3|  1   3   5    9    7   15   11    27   25    21  -> A003961
    4|  1   4   9   16   25   36   49    64   81   100  -> A000290
    5|  1   5   7   25   11   35   13   125   49    55  -> A357852
    6|  1   6  15   36   35   90   77   216  225   210  -> A191002
    7|  1   7  11   49   13   77   17   343  121    91
    8|  1   8  27   64  125  216  343   512  729  1000  -> A000578
    9|  1   9  25   81   49  225  121   729  625   441
   10|  1  10  21  100   55  210   91  1000  441   550
From _Peter Munn_, Jun 24 2021: (Start)
The encoding, n, of polynomials, f(n), that is used for the table is further described in A206284. Examples of encoded polynomials:
   n      f(n)        n           f(n)
   1         0       16              4
   2         1       17            x^6
   3         x       21        x^3 + x
   4         2       25           2x^2
   5       x^2       27             3x
   6     x + 1       35      x^3 + x^2
   7       x^3       36         2x + 2
   8         3       49           2x^3
   9        2x       55      x^4 + x^2
  10   x^2 + 1       64              6
  11       x^4       77      x^4 + x^3
  12     x + 2       81             4x
  13       x^5       90   x^2 + 2x + 1
  15   x^2 + x       91      x^5 + x^3
(End)
		

Crossrefs

Row n: n=1: A000012, n=2: A000027, n=3: A003961, n=4: A000290, n=5: A357852, n=6: A191002, n=8: A000578.
Main diagonal: A297473.
Functions f satisfying f(T(n,k)) = f(n) * f(k): A001222, A048675 (and similarly, other rows of A104244), A195017.
Powers of k: k=3: A000040, k=4: A001146, k=5: A031368, k=6: A007188 (see also A066117), k=7: A031377, k=8: A023365, k=9: main diagonal of A329050.
Integers in the ideal of the related ring (see Jun 2021 comment) generated by S: S={3}: A005408, S={4}: A000290\{0}, S={4,3}: A003159, S={5}: A007310, S={5,4}: A339690, S={6}: A325698, S={6,4}: A028260, S={7}: A007775, S={8}: A000578\{0}, S={8,3}: A191257, S={8,6}: A332820, S={9}: A016754, S={10,4}: A340784, S={11}: A008364, S={12,8}: A145784, S={13}: A008365, S={15,4}: A345452, S={15,9}: A046337, S={16}: A000583\{0}, S={17}: A008366.
Equivalent sequence for polynomial composition: A326376.
Related binary operations: A003991, A306697/A059896, A329329/A059897.

Programs

  • PARI
    T(n,k) = my (f=factor(n), p=apply(primepi, f[, 1]~), g=factor(k), q=apply(primepi, g[, 1]~)); prod (i=1, #p, prod(j=1, #q, prime(p[i]+q[j]-1)^(f[i, 2]*g[j, 2])))

Formula

T is completely multiplicative in both parameters:
- for any n > 0
- and k > 0 with prime factorization Prod_{i > 0} prime(i)^e_i:
- T(prime(n), k) = T(k, prime(n)) = Prod_{i > 0} prime(n + i - 1)^e_i.
For any m > 0, n > 0 and k > 0:
- T(n, k) = T(k, n) (T is commutative),
- T(m, T(n, k)) = T(T(m, n), k) (T is associative),
- T(n, 1) = 1 (1 is an absorbing element for T),
- T(n, 2) = n (2 is an identity element for T),
- T(n, 2^i) = n^i for any i >= 0,
- T(n, 4) = n^2 (A000290),
- T(n, 8) = n^3 (A000578),
- T(n, 3) = A003961(n),
- T(n, 3^i) = A003961(n)^i for any i >= 0,
- T(n, 6) = A191002(n),
- A001221(T(n, k)) <= A001221(n) * A001221(k),
- A001222(T(n, k)) = A001222(n) * A001222(k),
- A055396(T(n, k)) = A055396(n) + A055396(k) - 1 when n > 1 and k > 1,
- A061395(T(n, k)) = A061395(n) + A061395(k) - 1 when n > 1 and k > 1,
- T(A000040(n), A000040(k)) = A000040(n + k - 1),
- T(A000040(n)^i, A000040(k)^j) = A000040(n + k - 1)^(i * j) for any i >= 0 and j >= 0.
From Peter Munn, Mar 13 2020 and Apr 20 2021: (Start)
T(A329050(i_1, j_1), A329050(i_2, j_2)) = A329050(i_1+i_2, j_1+j_2).
T(n, m*k) = T(n, m) * T(n, k); T(n*m, k) = T(n, k) * T(m, k) (T distributes over multiplication).
A104244(m, T(n, k)) = A104244(m, n) * A104244(m, k).
For example, for m = 2, the above formula is equivalent to A048675(T(n, k)) = A048675(n) * A048675(k).
A195017(T(n, k)) = A195017(n) * A195017(k).
A248663(T(n, k)) = A048720(A248663(n), A248663(k)).
T(n, k) = A306697(n, k) if and only if T(n, k) = A329329(n, k).
A007913(T(n, k)) = A007913(T(A007913(n), A007913(k))) = A007913(A329329(n, k)).
(End)

Extensions

New name from Peter Munn, Jul 17 2021
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