Kieren MacMillan - cp's OEIS frontend

User: Kieren MacMillan

Kieren MacMillan has authored 3 sequences.

A230312 Squares which cannot be written as the sum of a smaller nonzero square and twice a triangular number.

1, 4, 9, 25, 49, 64, 100, 144, 169, 324, 400, 729, 784, 1089, 1369, 1764, 2025, 2209, 3025, 3364, 3600, 3844, 3969, 4225, 4489, 5329, 5625, 6084, 6400, 7225, 7744, 8100, 8464, 10404, 10609, 11025, 12544, 13225, 13924, 14400, 15625, 16384, 16900

Offset: 1

Kieren MacMillan, Dec 20 2013

The conjecture a(n) = A001912(n)^2 (mentioned in the formula part) is easy. In fact, any prime divisor of 4*n^2 + 1 is congruent to 1 modulo 4 and hence it can be written as a sum of two squares. Thus 4*n^2 + 1 = (2*n)^2 + 1^2 is composite if and only if it can be written as a sum of two squares in at least two ways. So the conjecture follows immediately. - Zhi-Wei Sun, Feb 23 2014

Positive squares that are the sum of two triangular numbers in exactly one way. Note that each positive square is the sum of two consecutive triangular numbers since A000217(n) + A000217(n+1) = n*(n+1)/2 + (n+1)*(n+2)/2 = (n+1)^2. - Altug Alkan, Jul 06 2016

			16 is not in the sequence because it can be expressed as 2^2 + 2 * 6.
But there is no such expression for 25 and hence it is in the sequence.

Cf. A001912.

A230312 = Reap[For[k = 1, k < 200, k++, n = k^2; If[Reduce[a > 0 && b > 0 && n == a^2 + b * (b + 1), {a, b}, Integers] == False, Sow[n]]]][[2, 1]] (* Jean-François Alcover, Dec 03 2014 *)

lista(nn) = for(n=1, nn, if(isprime(4*n^2+1), print1(n^2, ", "))); \\ Altug Alkan, Jul 06 2016

Conjecture: a(n) = A001912(n)^2, that is, squares of numbers n such that 4n^2 + 1 is prime. - Alonso del Arte, Dec 20 2013

A191975 Least common multiple of all p-1, where prime p divides the n-th primary pseudoperfect number A054377(n).

Original entry on oeis.org

1, 2, 6, 42, 330, 235290, 310800, 1863851053628494074457830

Offset: 1

Kieren MacMillan, Jun 20 2011

a(n) is a factor of any exponent k > 0 such that 1^k + 2^k + ... + p^k == 1 (mod p), where p = A054377(n).

			A054377(3) = 42 = 2*3*7, so a(3) = lcm(2-1, 3-1, 7-1) = lcm(1,2,6) = 6.

J. Sondow and K. MacMillan, Reducing the Erdős-Moser equation 1^n + 2^n + ... + k^n = (k+1)^n modulo k and k^2, Integers 11 (2011), #A34.

Cf. A054377.

a(n) = lcm(p-1 : prime p | A054377(n)).

A147297 Primes of the form (2k)^2 + 3(2k + 1)^2.

Original entry on oeis.org

31, 307, 463, 1123, 1723, 3307, 4831, 6007, 8011, 10303, 11131, 13807, 20023, 23563, 26083, 30103, 35911, 43891, 60271, 86143, 95791, 108571, 127807, 136531, 145543, 164431, 205663, 239611, 276151, 284623, 288907, 366631, 371491, 386263, 459007

Offset: 1

Kieren MacMillan, Nov 05 2008

nonn

First thirteen terms are a subset of A073337, A002383 and A085104.

[ a: n in [1..900] | IsPrime(a) where a is (2*n)^2 + 3*(2*n+1)^2] // Vincenzo Librandi, Nov 25 2010

makelist((2*k)^2+3*(2*k+1)^2,k,1,100)$ sublist(%,primep);

More terms from Vincenzo Librandi, Apr 28 2010

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User: Kieren MacMillan

A230312 Squares which cannot be written as the sum of a smaller nonzero square and twice a triangular number.

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A191975 Least common multiple of all p-1, where prime p divides the n-th primary pseudoperfect number A054377(n).

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Formula

A147297 Primes of the form (2k)^2 + 3(2k + 1)^2.

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