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.

Showing 1-10 of 32 results. Next

A271576 Integers whose square is of the form x^2 + y^4, with x,y > 0 (see A111925).

Original entry on oeis.org

5, 15, 20, 34, 39, 41, 45, 60, 65, 80, 85, 111, 125, 135, 136, 145, 150, 156, 164, 175, 180, 194, 219, 240, 245, 255, 260, 265, 306, 313, 320, 325, 340, 351, 353, 369, 371, 375, 405, 410, 444, 445, 455, 500, 505, 514, 540, 544, 580, 585, 600, 605, 609, 624, 629, 656, 671, 674, 689
Offset: 1

Views

Author

Michel Marcus, Apr 10 2016

Keywords

Comments

If z is in this sequence, z*k^2 is, for k > 0. Also note that since there are no fourth powers in A111925, there are no squares in this sequence. - Altug Alkan, Apr 10 2016
From Karl-Heinz Hofmann, Oct 22 2021: (Start)
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. (End)
From Jon E. Schoenfield, Nov 15 2021: (Start)
Apparently, the vast majority of the terms of this sequence can be expressed as x^2 + y^4 with x,y > 0 in only one way (A345645 lists those terms), but some can be so expressed in exactly two, three, four, five, six, or more ways. Among the first 3976926961 terms of this sequence,
3948648229 are in A345645 (exactly 1 way),
25415062 are in A345700 (exactly 2 ways),
2697713 are in A345968 (exactly 3 ways),
161543 are in A346110 (exactly 4 ways),
3989 are in A348655 (exactly 5 ways),
424 are in A349324 (exactly 6 ways),
and just 1 -- a(3976926961) = 2474052064291275 = A346115(7) -- is a number whose square can be written as x^2 + y^4 with x,y > 0 in exactly 7 ways. The ratios of successive counts above, i.e., 3948648229/25415062, 25415062/2697713, 2697713/161543, 161543/3989, 3989/424, and 424/1, are approximately 155.4, 9.4, 16.7, 40.5, 9.4, and 424.0. What is it that (over the interval [1, 2474052064291275], at least) makes numbers whose squares can be written as x^2 + y^4 with x,y > 0 in more than 6 ways so much rarer than those that can be written thus in exactly 6 ways? (End)

Examples

			5^2 = 25 = 9 + 16 = 3^2 + 2^4, so 5 is a term.
		

Crossrefs

Cf. A111925, A271577, A345645 (in exactly 1 way).
Cf. A345700 (in exactly 2 ways), A345968 (in exactly 3 ways).
Cf. A346110 (in exactly 4 ways), A349324 (in exactly 6 ways), A346115 (the least solutions).
Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

Programs

  • Mathematica
    Select[Range@ 200, Resolve[Exists[{x, y}, Reduce[#^2 == x^2 + y^4, {x, y}, Integers], And[x > 0, y > 0]]] &] (* Michael De Vlieger, Apr 10 2016 *)
  • PARI
    isok(n) = n = n^2; for (k=1, n-1, if (issquare(k) && (p=ispower(n-k, 4)), return (1)))
    
  • PARI
    is(n)=my(n2=n^2); for(b=sqrtnint(2*n-2,4)+1, sqrtint(n-1), if(issquare(n2-b^4), return(1))); 0 \\ Charles R Greathouse IV, Nov 16 2021

A271577 Powers, squares excepted, of the form x^2+y^4, with x,y > 0 (see A111925).

Original entry on oeis.org

32, 512, 2197, 3125, 8000, 8192, 15625, 64000, 100000, 131072, 1419857, 1953125, 2097152, 3200000, 7189057, 8000000, 8998912, 9765625, 11390625, 11881376, 30664297, 32768000, 33554432, 35287552, 62748517, 64000000, 69343957, 115856201, 125000000, 221445125, 262144000, 272097792
Offset: 1

Views

Author

Michel Marcus, Apr 10 2016

Keywords

Examples

			32 = 2^5 = 16 + 16 = 2^2 + 2^4, so 32 is a term.
		

Crossrefs

Programs

  • PARI
    isok(n) = {if ((p=ispower(n)) && (p != 2), for (k=1, n-1, if (issquare(k) && (p=ispower(n-k, 4)), return (1));););}

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

Views

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

A002523 a(n) = n^4 + 1.

Original entry on oeis.org

1, 2, 17, 82, 257, 626, 1297, 2402, 4097, 6562, 10001, 14642, 20737, 28562, 38417, 50626, 65537, 83522, 104977, 130322, 160001, 194482, 234257, 279842, 331777, 390626, 456977, 531442, 614657, 707282, 810001, 923522, 1048577, 1185922, 1336337, 1500626, 1679617
Offset: 0

Views

Author

Keywords

Comments

a(n) = Phi_8(n), where Phi_k is the k-th cyclotomic polynomial.
All odd prime factors of a(n) are congruent to 1 modulo 8. - Nick Hobson, Jan 14 2007
Lee and Murty, p. 685: "In spite of these remarkable advances, we are still unable to determine if n^4 + 1 is infinitely often a squarefree number". - Jonathan Vos Post, Sep 18 2007
Since a(n)*a(m) = (n^4+1)*(m^4+1) = ((n*m)^2-1)^2 + (n^2+m^2)^2, a(n)*a(m) is obvious member of A000404 for n*m > 1. Additionally, if m and n are the legs of a Pythagorean triple, then a(m)*a(n) is the member of A111925. - Altug Alkan, Apr 08 2016

References

  • M. Mabkhout, "Minoration de P(x^4+1)", Rend. Sem. Fac. Sci. Univ. Cagliari 63 (2) (1993), 135-148.

Crossrefs

Programs

Formula

From R. J. Mathar, Apr 28 2008: (Start)
O.g.f.: (1 - 3*x + 17*x^2 + 7*x^3 + 2*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)
Sum_{n>=0} 1/a(n) = 1/2 + Pi * (sinh(sqrt(2)*Pi) + sin(sqrt(2)*Pi)) / (2*sqrt(2) * (cosh(sqrt(2)*Pi) - cos(sqrt(2)*Pi))) = 1.578477579667136838318... . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} (-1)^n/a(n) = 1/2 - Pi * (cos(Pi/sqrt(2)) * sinh(Pi/sqrt(2)) + cosh(Pi/sqrt(2)) * sin(Pi/sqrt(2))) / (sqrt(2) * (cos(sqrt(2)*Pi) - cosh(sqrt(2)*Pi))) = 0.54942814871987317922929... . - Vaclav Kotesovec, Feb 14 2015
Product_{n>=1} (1 - 1/a(n)) = 2*Pi^2/(cosh(sqrt(2)*Pi) - cos(sqrt(2)*Pi)). - Amiram Eldar, Jan 26 2024

A022549 Sum of a square and a nonnegative cube.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 12, 16, 17, 24, 25, 26, 27, 28, 31, 33, 36, 37, 43, 44, 49, 50, 52, 57, 63, 64, 65, 68, 72, 73, 76, 80, 81, 82, 89, 91, 100, 101, 108, 113, 121, 122, 125, 126, 127, 128, 129, 134, 141, 144, 145
Offset: 1

Views

Author

Keywords

Comments

It appears that there are no modular constraints on this sequence; i.e., every residue class of every integer has representatives here. - Franklin T. Adams-Watters, Dec 03 2009
A045634(a(n)) > 0. - Reinhard Zumkeller, Jul 17 2010

Crossrefs

Complement of A022550; A002760 and A179509 are subsequences.

Programs

  • Mathematica
    q=30; imax=q^2; Select[Union[Flatten[Table[x^2+y^3, {y,0,q^(2/3)}, {x,0,q}]]], #<=imax&] (* Vladimir Joseph Stephan Orlovsky, Apr 20 2011 *)
  • PARI
    is(n)=for(k=0,sqrtnint(n,3), if(issquare(n-k^3), return(1))); 0 \\ Charles R Greathouse IV, Aug 24 2020
    
  • PARI
    list(lim)=my(v=List(),t); for(k=0,sqrtnint(lim\=1,3), t=k^3; for(n=0,sqrtint(lim-t), listput(v,t+n^2))); Set(v) \\ Charles R Greathouse IV, Aug 24 2020

A100291 Numbers of the form a^4 + b^3 with a, b > 0.

Original entry on oeis.org

2, 9, 17, 24, 28, 43, 65, 80, 82, 89, 108, 126, 141, 145, 206, 217, 232, 257, 264, 283, 297, 320, 344, 359, 381, 424, 472, 513, 528, 593, 599, 626, 633, 652, 689, 730, 745, 750, 768, 810, 841, 968, 985, 1001, 1016, 1081, 1137, 1256, 1297, 1304, 1323, 1332
Offset: 1

Views

Author

T. D. Noe, Nov 18 2004

Keywords

Crossrefs

Cf. A100271 (primes of the form a^4 + b^3).
Cf. A055394 (a^2 + b^3: contains this as subsequence), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4), A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).
Roots of 5th powers are listed in A300565 (z^5 = x^3 + y^4); see also A300564 (z^4 = x^2 + y^3) and A242183, A300566 (z^6 = x^4 + y^5), A300567 (z^7 = x^6 + y^5), A302174.

Programs

  • Mathematica
    lst={}; Do[p=a^4+b^3; If[p<2000, AppendTo[lst, p]], {a, 64}, {b, 256}]; Union[lst]
    With[{nn=20},Select[Union[#[[1]]^4+#[[2]]^3&/@Tuples[Range[20],2]],#<= nn^3+1&]] (* Harvey P. Dale, May 27 2020 *)
  • PARI
    is(n)=for(a=1, sqrtnint(n-1, 4), ispower(n-a^4, 3) && return(a)) \\ Returns a > 0 if n is in the sequence, or 0 otherwise. - M. F. Hasler, Apr 25 2018
    
  • PARI
    list(lim)=my(v=List());for(b=1,sqrtnint(lim\=1,3), my(b3=b^3); for(a=1,sqrtnint(lim-b3,4), listput(v,a^4+b3))); Set(v) \\ Charles R Greathouse IV, Jul 26 2021

Extensions

Edited by M. F. Hasler, Apr 25 2018

A100292 Numbers of the form a^5 + b^2 with a, b > 0.

Original entry on oeis.org

2, 5, 10, 17, 26, 33, 36, 37, 41, 48, 50, 57, 65, 68, 81, 82, 96, 101, 113, 122, 132, 145, 153, 170, 176, 197, 201, 226, 228, 244, 247, 252, 257, 259, 268, 279, 288, 290, 292, 307, 321, 324, 325, 343, 356, 362, 364, 387, 393, 401, 412, 432, 439, 442, 468, 473
Offset: 1

Views

Author

T. D. Noe, Nov 18 2004

Keywords

Crossrefs

Cf. A100272 (primes of the form a^5 + b^2).
Cf. A000404 (a^2 + b^2), A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4), A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).

Programs

  • Mathematica
    lst={}; Do[p=a^5+b^2; If[p<1000, AppendTo[lst, p]], {a, 16}, {b, 1024}]; Union[lst]
  • PARI
    is(n, m=5)=for(a=1, sqrtnint(n-1, m), issquare(n-a^m) && return(a)) \\ M. F. Hasler, Apr 25 2018

A100294 Numbers of the form a^5 + b^4 with a, b > 0.

Original entry on oeis.org

2, 17, 33, 48, 82, 113, 244, 257, 259, 288, 324, 499, 626, 657, 868, 1025, 1040, 1105, 1280, 1297, 1328, 1539, 1649, 2320, 2402, 2433, 2644, 3126, 3141, 3206, 3381, 3425, 3750, 4097, 4128, 4339, 4421, 5120, 5526, 6562, 6593, 6804, 7221, 7585, 7777, 7792
Offset: 1

Views

Author

T. D. Noe, Nov 18 2004

Keywords

Comments

In view of computing A300566, it would be interesting to have an efficient way to check whether a given (large) n is in this sequence. - M. F. Hasler, Apr 25 2018

Crossrefs

Cf. A100274 (primes of the form a^5 + b^4).
Subsequence of A100292 (a^5 + b^2); see also A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100293 (a^5 + b^3), A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).
Roots of 6th powers are listed in A300566 (z such that z^6 = x^5 + y^4 for some x, y >= 1); see also A300564 (z^4 = x^3 + y^2) and A242183, A300565 (z^5 = x^4 + y^3), A300567 (z^7 = x^6 + y^5), A302174.

Programs

  • Mathematica
    lst={}; Do[p=a^5+b^4; If[p<15000, AppendTo[lst, p]], {a, 16}, {b, 32}]; Union[lst]
  • PARI
    A100294_vec(L=10^6, k=4, m=5, S=List())={for(a=1, sqrtnint(L-1, m), for(b=1, sqrtnint(L-a^m, k), listput(S, a^m+b^k))); Set(S)} \\ all terms up to limit L. - M. F. Hasler, Apr 25 2018
    
  • PARI
    is(n, k=4, m=5)=for(a=1, sqrtnint(n-1, m), ispower(n-a^m,k) && return(a)) \\ Returns a > 0 if n is in the sequence, or 0 otherwise. - M. F. Hasler, Apr 25 2018

A303375 Numbers of the form a^5 + b^6, with integers a, b > 0.

Original entry on oeis.org

2, 33, 65, 96, 244, 307, 730, 761, 972, 1025, 1088, 1753, 3126, 3189, 3854, 4097, 4128, 4339, 5120, 7221, 7777, 7840, 8505, 11872, 15626, 15657, 15868, 16649, 16808, 16871, 17536, 18750, 20903, 23401, 32432, 32769, 32832, 33497, 36864, 46657, 46688, 46899, 47680, 48393
Offset: 1

Views

Author

M. F. Hasler, Apr 22 2018

Keywords

Comments

Although it is easy to produce many terms of this sequence, it is nontrivial to check whether a very large number is of this form.
This sequence is among others motivated by the hard-to-compute sequence A300567 = numbers z such that z^7 = x^5 + y^6 for some x, y >= 1.

Crossrefs

Cf. A000404 (a^2 + b^2), A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4).
Cf. A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303374 (a^4 + b^6).
See also A300567: numbers z such that z^7 = x^5 + y^6 for some x, y >= 1.

Programs

  • PARI
    is(n,k=5,m=6)=for(b=1,sqrtnint(n-1,m),ispower(n-b^m,n)&&return(b)) \\ Returns b > 0 if n is in the sequence, else 0.
    A303375_vec(L=10^5,k=5,m=6,S=List())={for(a=1,sqrtnint(L-1,m),for(b=1,sqrtnint(L-a^m,k), listput(S,a^m+b^k)));Set(S)} \\ all terms up to limit L

Formula

a(n) >> n^(30/11). Probably this is the correct asymptotic order. - Charles R Greathouse IV, Jan 23 2025

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

Views

Author

Karl-Heinz Hofmann, Jul 05 2021

Keywords

Comments

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

Examples

			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).
		

Crossrefs

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)).

Programs

  • Python
    # See Hofmann link.
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