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 1-3 of 3 results.

A336841 Prime-shifted analog of A094471: a(n) = A336845(n) - A003973(n).

Original entry on oeis.org

0, 2, 4, 14, 6, 36, 10, 68, 44, 52, 12, 192, 16, 84, 92, 284, 18, 326, 22, 274, 148, 100, 28, 840, 90, 132, 344, 438, 30, 648, 36, 1094, 176, 148, 212, 1622, 40, 180, 232, 1192, 42, 1032, 46, 520, 802, 228, 52, 3324, 230, 654, 260, 684, 58, 2376, 252, 1896, 316, 244, 60, 3156, 66, 292, 1278, 4010, 332, 1224, 70, 766
Offset: 1

Views

Author

Antti Karttunen, Aug 06 2020

Keywords

Comments

All terms are even because A003973 and A336845 match parity-wise. Also in the sum formulas, only even terms are summed (only one of which is zero).

Links

Crossrefs

Cf. A000005, A000203, A003961, A048673, A094471, A336837, A336839, A336840, A336845, A336856.
Cf. A336846 [= gcd(a(n), A003973(n))].
Twice the terms of A336854.

Programs

Formula

a(n) = A336845(n) - A003973(n) = (A000005(n)*A003961(n)) - A000203(A003961(n)).
a(n) = A094471(A003961(n)).
a(n) = Sum_{d|n} (A003961(n)-A003961(d)) = Sum_{d|A003961(n)} (A003961(n)-d).
a(n) = 2*A336854(n) = 2*Sum_{d|n} (A048673(n)-A048673(d)).
a(n) = ((A003961(n)+1)*A000005(n)) - 2*A336840(n).
a(n) = 2 * ((A000005(n)*A048673(n)) - A336840(n)).
a(n) = A000005(n) * (A336837(n)/A336839(n)) = A336837(n) * A336856(n).

A336839 Denominator of the arithmetic mean of the divisors of A003961(n).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Aug 07 2020

Keywords

Comments

Also denominator of A336841(n) / A000005(n).
All terms are odd because A336932(n) = A007814(A003973(n)) >= A295664(n) for all n.

Links

Crossrefs

Cf. A000005, A000225, A003961, A003973, A007814, A052940, A057021, A295664, A336840, A336841, A336856, A336931, A336932.
Cf. A336918 (positions of 1's), A336919 (of terms > 1).
Cf. A336837 and A336838 (numerators).

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A336839(n) = denominator(sigma(A003961(n))/numdiv(n));

Formula

a(n) = denominator(A003973(n)/A000005(n)).
a(n) = d(n)/A336856(n) = d(n)/gcd(d(n),A003973(n)) = d(n)/gcd(d(n),A336841(n)), where d(n) is the number of divisors of n, A000005(n).
a(n) = A057021(A003961(n)).
For all primes p, and e >= 0, a(A000225(e)) = a(p^((2^e) - 1)) = 1. [See A336856]
It seems that for all odd primes p, and with the exponents e=5, 11, 17 or 23 (at least these), a(p^e) = 1.
It seems that a(27^((2^n)-1)) = A052940(n-1) for all n >= 1.

A336856 Prime-shifted analog of gcd(d(n), sigma(n)): a(n) = gcd(A000005(n), A003973(n)).

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 2, 4, 1, 4, 2, 6, 2, 4, 4, 1, 2, 2, 2, 2, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 2, 4, 4, 4, 1, 2, 4, 4, 8, 2, 8, 2, 2, 2, 4, 2, 2, 1, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 1, 4, 8, 2, 2, 4, 8, 2, 4, 2, 4, 6, 6, 4, 8, 2, 2, 1, 4, 2, 12, 4, 4, 4, 8, 2, 4, 4, 6, 4, 4, 4, 12, 2, 2, 2, 3, 2, 8, 2, 8, 8
Offset: 1

Views

Author

Antti Karttunen, Aug 12 2020

Keywords

Links

Crossrefs

Cf. A000005, A003961, A003973, A009205, A336837, A336838, A336839, A336840, A336841.

Programs

  • PARI
    A003973(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); sigma(factorback(f)); };
A336856(n) = gcd(numdiv(n), A003973(n));

Formula

a(n) = A009205(A003961(n)).
a(n) = gcd(A000005(n), A003973(n)) = gcd(A000005(n), A336841(n)).
a(n) = gcd(A000005(n), 2*A336840(n)).
a(n) = A003973(n) / A336838(n) = A000005(n) / A336839(n).
For n > 1, a(n) = A336841(n) / A336837(n).
For all primes p, and n >= 0, a(p^((2^n)-1)) = 2^n.
Showing 1-3 of 3 results.

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