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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A269928 Integers n that belong to more Pythagorean triples than preceding integers.

Original entry on oeis.org

1, 3, 5, 12, 15, 24, 40, 48, 60, 120, 240, 360, 420, 720, 840, 1560, 1680, 2520, 3360, 5040, 8400, 9240, 10920, 15120, 18480, 21840, 27720, 32760, 36960, 43680, 55440, 65520, 109200, 110880, 120120, 166320, 196560, 221760, 240240, 360360, 480480, 720720, 1113840
Offset: 1

Views

Author

Michel Marcus, Mar 08 2016

Keywords

Comments

Called optimal P-numbers in Tripathi link.

Crossrefs

Programs

  • PARI
    nbpt(n) = {oddn = n/(2^valuation(n, 2)); f = factor(oddn); for (k=1, #f~, if ((f[k,1] % 4) != 1, f[k,2] = 0);); n1 = factorback(f); if (n % 2, (numdiv(n^2)+numdiv(n1^2))/2 -1, (numdiv((n/2)^2)+numdiv(n1^2))/2 -1);}
    lista(nn) = {last = -1; for (n=1, nn, if ((new = nbpt(n)) > last, print1(n, ", "); last = new;););}

A354048 a(n) is the largest number of distinct integer-sided right triangles in which some n-digit number can appear as the length of a side.

Original entry on oeis.org

2, 14, 68, 203, 476, 1421, 3293, 7910, 20060, 39509, 89324, 206711, 442907, 803924, 1722464, 3198608, 6820523, 13434254, 27901259, 50222267
Offset: 1

Views

Author

Zhining Yang, Jun 26 2022

Keywords

Examples

			a(2)=14 because there exist 14 distinct integer-sided right triangles with the 2-digit number 60 as the length of a side, i.e., (11,60,61), (25,60,65), (32,60,68), (36,48,60), (45,60,75), (60,63,87), (60,80,100), (60,91,109), (60,144,156), (60,175,185), (60,221,229), (60,297,303), (60,448,452), and (60,899,901), and no 2-digit number is the length of a side of more than 14 distinct integer-sided right triangles.
		

Crossrefs

Programs

  • Python
    from sympy import factorint
    def s(n):
        f=factorint(n)
        d, q=(list(f.keys()), list(f.values()))
        (a, b, c, x)=(0, 1, 1, 0)
        if(d[0]==2):
            a, x=(0, 1)
            if q[0]>1:
                 a=q[0]-1
        for p in range(x, len(d)):
            b*=(1+2*q[p])
            if d[p]%4==1:
                c*=(1+2*q[p])
        return((b-1)//2+a*b+(c-1)//2)
    def a(n):
        max=0
        for i in range(1+10**(n-1), 10**n):
            if s(i)>max:
                k,max=(i,s(i))
        return(n,[k,max])
    for i in range(1,6):
        print (a(i))
    # (thanks to Zhao Hui Du for help in the derivation of this function)
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