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 1-6 of 6 results.

A145299 Smallest k such that k^2+1 is divisible by A002144(n)^6.

Original entry on oeis.org

1068, 1999509, 390112, 253879357, 756360062, 2363588163, 5041394261, 9435321777, 41865466758, 102666405913, 197177418061, 316411915250, 171829799914, 625667121807, 182312430890, 1095001339019, 6390289199260
Offset: 1

Views

Author

Klaus Brockhaus, Oct 17 2008

Keywords

Examples

			a(1) = 1068 since A002144(1) = 5, 1068^2+1 = 1140625 = 5^6*73 and for no k < 1068 does 5^6 divide k^2+1. a(11) = 197177418061 since A002144(11) = 97, 197177418061^2+1 = 38878934193202368999722 = 2*97^6*23337479509 and for no k < 197177418061 does 97^6 divide k^2+1.
		

Crossrefs

Cf. A002144 (primes of form 4n+1), A002313 (-1 is a square mod p), A059321, A145296, A145297, A145298.

Programs

  • PARI
    { e=6; forprime(p=2, 1000, if(p%4==1, k=lift(sqrt(-1+O(p^e))); if(k>p^e/2,k=p^e-k); print1(k, ", "))) }
    
  • Python
    from itertools import islice
    from sympy import nextprime, sqrt_mod_iter
    def A145299_gen(): # generator of terms
        p = 1
        while (p:=nextprime(p)):
            if p&3==1:
                yield min(sqrt_mod_iter(-1,p**6))
    A145299_list = list(islice(A145299_gen(),20)) # Chai Wah Wu, May 04 2024

Extensions

More terms and efficient PARI program from. - Max Alekseyev, Oct 28 2008

A145297 Smallest k such that k^2+1 is divisible by A002144(n)^4.

Original entry on oeis.org

182, 239, 27493, 34522, 800982, 1251967, 623098, 6304056, 6459524, 20099637, 22709274, 35764191, 40317977, 54397650, 166206108, 187800003, 165728858, 152475014, 282599844, 312923750, 154613663, 485200742, 912190662, 548850444
Offset: 1

Views

Author

Klaus Brockhaus, Oct 11 2008

Keywords

Examples

			a(1) = 182 since A002144(1) = 5, 182^2+1 = 33125 = 5^4*53 and for no k < 182 does 5^4 divide k^2+1.
		

Crossrefs

Cf. A002144 (primes of form 4n+1), A002313 (-1 is a square mod p), A059321, A145296, A145298, A145299.

Programs

  • PARI
    {e=4; forprime(p=2, 250, if(p%4==1, q=p^e; m=q; while(!ispower(m-1,2,&n), m=m+q); print1(n, ",")))}
    
  • Python
    from itertools import islice
    from sympy import nextprime, sqrt_mod_iter
    def A145297_gen(): # generator of terms
        p = 1
        while (p:=nextprime(p)):
            if p&3==1:
                yield min(sqrt_mod_iter(-1,p**4))
    A145297_list = list(islice(A145297_gen(),20)) # Chai Wah Wu, May 04 2024

A145298 Smallest k such that k^2+1 is divisible by A002144(n)^5.

Original entry on oeis.org

1068, 143044, 390112, 7745569, 6423465, 46464143, 23048345, 144762466, 404034898, 2153335831, 331407850, 1108900220, 2581164875, 760839155, 10734466938, 6595297216, 773302059, 61063137802, 31915893786, 112699451831
Offset: 1

Views

Author

Klaus Brockhaus, Oct 14 2008

Keywords

Examples

			a(4) = 7745569 since A002144(4) = 29, 7745569^2+1 = 59993839133762 = 2*29^5*97*15077 and for no k < 7745569 does 29^5 divide k^2+1.
		

Crossrefs

Cf. A002144 (primes of form 4n+1), A002313 (-1 is a square mod p), A059321, A145296, A145297, A145299.

Programs

  • PARI
    {e=5; forprime(p=2, 200, if(p%4==1, q=p^e; m=q; while(!ispower(m-1,2,&n), m=m+q); print1(n, ",")))}
    
  • Python
    from itertools import islice
    from sympy import nextprime, sqrt_mod_iter
    def A145298_gen(): # generator of terms
        p = 1
        while (p:=nextprime(p)):
            if p&3==1:
                yield min(sqrt_mod_iter(-1,p**5))
    A145298_list = list(islice(A145298_gen(),20)) # Chai Wah Wu, May 04 2024

A145871 Smallest k such that k^2+1 is divisible by A002144(n)^7.

Original entry on oeis.org

32318, 6826318, 96940388, 7986582530, 24900904028, 92615568742, 416081467190, 988322434636, 3219884218827, 4867146503697, 26457926739667, 47023298541694, 26661771973542, 90980209992989, 257680081342861, 283410689912607
Offset: 1

Views

Author

Klaus Brockhaus, Oct 22 2008

Keywords

Examples

			a(2) = 6826318 since A002144(2) = 13, 6826318^2+1 = 46598617437125 = 5^3*13^7*13*457 and for no k < 6826318 does 13^7 divide k^2+1. a(4) = 7986582530 since A002144(4) = 29, 7986582530^2+1 = 63785500508501200901 = 29^7*197*409*45893 and for no k < 7986582530 does 29^7 divide k^2+1.
		

Crossrefs

Cf. A002144 (primes of form 4n+1), A002313 (-1 is a square mod p), A059321, A145296, A145297, A145298, A145299, A145872, A145873.

Programs

  • PARI
    {e=7; forprime(p=2, 40, if(p%4==1, q=p^e; m=q; while(!issquare(m-1, &n), m=m+q); print1(n, ",")))}

Extensions

More terms from Klaus Brockhaus, Nov 12 2008

A145872 Smallest k such that k^2+1 is divisible by A002144(n)^8.

Original entry on oeis.org

110443, 6826318, 3379649772, 61012922706, 1019349744435, 287369842623, 11331029931180, 71294762793847, 239822883201307, 923990886302412, 2369608176604944, 3156215819652023, 521749964271465, 2026364722410364
Offset: 1

Views

Author

Klaus Brockhaus, Oct 22 2008

Keywords

Examples

			a(1) = 110443 since A002144(1) = 5, 110443^2+1 = 12197656250 = 2*5^8*13*1201 and for no k < 110443 does 5^8 divide k^2+1. a(3) = 3379649772 since A002144(3) = 17, 3379649772^2+1 = 11422032581379651985 = 5*13*17^8*97*259697 and for no k < 3379649772 does 17^8 divide k^2+1.
		

Crossrefs

Cf. A002144 (primes of form 4n+1), A002313 (-1 is a square mod p), A059321, A145296, A145297, A145298, A145299, A145871, A145873.

Programs

  • PARI
    {e=8; forprime(p=2, 40, if(p%4==1, q=p^e; m=q; while(!issquare(m-1, &n), m=m+q); print1(n, ",")))}

Extensions

More terms from Klaus Brockhaus, Nov 12 2008

A145873 Smallest k such that k^2+1 is divisible by A002144(n)^9.

Original entry on oeis.org

280182, 822557039, 24306922095, 4563230639355, 15069267560119, 112076323050317, 50928660480181, 3138611770750343, 9110883894036198, 50251663587824641, 76004727767164666, 310872228812491206, 521749964271465
Offset: 1

Views

Author

Klaus Brockhaus, Oct 30 2008

Keywords

Examples

			a(1) = 280182 since A002144(1) = 5, 280182^2+1 = 78501953125 = 5^9*40193 and for no k < 280182 does 5^9 divide k^2+1. a(3) = 24306922095 since A002144(3) = 17, 24306922095^2+1 = 590826461732399189026 = 2*17^9*29*673*127637 and for no k < 24306922095 does 17^9 divide k^2+1.
		

Crossrefs

Cf. A002144 (primes of form 4n+1), A002313 (-1 is a square mod p), A059321, A145296, A145297, A145298, A145299, A145871, A145872.
Showing 1-6 of 6 results.