A232616 -id:A232616 - cp's OEIS frontend

A232398 Number of ways to write n = p + (2^k - k) + (2^m - m) with p prime and 0 < k <= m.

0, 0, 0, 1, 2, 2, 2, 2, 4, 2, 2, 2, 3, 2, 4, 3, 4, 2, 4, 4, 4, 2, 3, 3, 3, 4, 4, 1, 3, 4, 5, 3, 5, 4, 5, 4, 4, 1, 4, 3, 5, 3, 5, 4, 5, 4, 5, 3, 3, 4, 5, 2, 3, 3, 4, 4, 5, 3, 3, 4, 6, 4, 5, 3, 7, 5, 5, 3, 4, 6, 6, 4, 7, 4, 6, 6, 7, 3, 3, 4, 5, 5, 6, 2, 6, 5, 5, 4, 5, 5, 5, 5, 5, 1, 4, 6, 4, 2, 5, 6

Offset: 1

Zhi-Wei Sun, Nov 23 2013

nonn

Conjecture: a(n) > 0 for all n > 3.
This was motivated by A231201. We have verified the conjecture for n up to 2*10^8. It seems that a(n) = 1 for no odd n.
In contrast, R. Crocker proved that there are infinitely many positive odd numbers not of the form p + 2^k + 2^m with p prime and k, m > 0.
It seems that any integer n > 3 not equal to 1361802 can be written in the form p + (2^k + k) + (2^m + m), where p is a prime, and k and m are nonnegative integers.
On Dec 08 2013, Qing-Hu Hou finished checking the conjecture for n up to 10^10 and found no counterexamples. - Zhi-Wei Sun, Dec 08 2013

			a(11) = 2 since 11 = 5 + (2 - 1) + (2^3 - 3) = 7 + (2^2 - 2) + (2^2 - 2) with 5 and 7 prime.
a(28) = 1 since 28 = 11 + (2^3 - 3) + (2^4 - 4) with 11 prime.
a(94) = 1 since 94 = 31 + (2^3 - 3) + (2^6 - 6) with 31 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
R. Crocker, On the sum of a prime and two powers of two, Pacific J. Math. 36 (1971), 103-107.
Zhi-Wei Sun, On a^n + b*n modulo m, preprint, arXiv:1312.1166 [math.NT], 2013-2014.

Cf. A000040, A000079, A156695, A231201, A231725, A232616.

PQ[n_]:=n>0&&PrimeQ[n]
A232398[n_] := Sum[If[2^m - m < n && PQ[n - 2^m + m - 2^k + k], 1, 0], {m, Log[2, 2n]}, {k, m}]; Table[A232398[n], {n, 100}]

A232862 Least positive integer m <= n^2/2 + 3 such that the set {prime(k) - k: k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 3, 4, 11, 9, 8, 10, 15, 29, 13, 23, 22, 23, 37, 32, 28, 48, 44, 53, 41, 45, 67, 76, 117, 119, 91, 121, 88, 89, 101, 72, 88, 100, 143, 144, 185, 145, 104, 176, 141, 144, 175, 187, 213, 121, 255, 128, 129, 189, 243, 122, 267, 275, 242, 209, 205, 130, 153, 263, 335

Offset: 1

Zhi-Wei Sun, Dec 01 2013

nonn

Conjecture: a(n) > 0 for all n > 0.

Note that a(4) = 11 = 4^2/2 + 3.

			a(3) = 4 since prime(1) - 1 = prime(2) - 2 = 1, prime(3) - 3 = 2, prime(4) - 4 = 3, and {1,2,3} is a complete system of residues modulo 3.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from Zhi-Wei Sun)
Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.

Cf. A014689, A232616, A232463, A232548, A232861.

  L[m_,n_]:=Length[Union[Table[Mod[Prime[k]-k,n],{k,1,m}]]]
Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+3}];
Print[n," ",0];Label[aa];Continue,{n,1,60}]

A232894 Least positive integer m such that {Catalan(k) - k: k = 1, ..., m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 5, 4, 11, 16, 13, 31, 27, 18, 22, 34, 52, 45, 45, 31, 112, 57, 73, 113, 99, 64, 77, 114, 215, 134, 106, 89, 99, 127, 209, 161, 239, 135, 178, 96, 207, 185, 172, 157, 231, 174, 195, 309, 115, 274, 309, 386, 239, 200, 336, 188, 199, 181, 181, 116, 311, 229, 290, 663, 239

Offset: 1

Zhi-Wei Sun, Dec 02 2013

nonn

Conjecture: (i) Let n be any positive integer. Then 0 < a(n) <= n^2/2 + 7. Also, {Catalan(k) + k: k = 1, ..., [n^2/2] + 23} contains a complete system of residues modulo n, where [.] is the floor function.

(ii) For any integer n > 3, neither Catalan(n) - n nor Catalan(n) + n has the form x^m with m > 1 and x > 1.

			a(2) = 5 since Catalan(k) - k is even for each k = 1, 2, 3, 4, and Catalan(5) - 5 = 37 is odd.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..200 from Zhi-Wei Sun)

Cf. A000108, A232616, A232862, A232898.

L[m_,n_]:=Length[Union[Table[Mod[CatalanNumber[k]-k,n],{k,1,m}]]]
Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+7}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]

A232898 Least positive integer m such that {C(2k,k) + k: k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 2, 7, 5, 10, 12, 9, 24, 31, 22, 59, 25, 27, 30, 42, 56, 123, 66, 57, 72, 84, 78, 73, 132, 136, 57, 99, 80, 129, 211, 170, 226, 121, 170, 126, 129, 238, 218, 157, 132, 348, 198, 388, 103, 171, 166, 247, 181, 205, 352, 194, 136, 430, 226, 117, 224, 237, 292, 364, 241

Offset: 1

Zhi-Wei Sun, Dec 02 2013

nonn

Conjecture: (i) Let n be any positive integer. Then 0 < a(n) <= n^2/2 + 3. Also, {C(2k,k) - k: k = 1, ..., [n^2/2] + 15} contains a complete system of residues modulo n, where [.] is the floor function.

(ii) For any integer n > 2, neither C(2n,n) + n nor C(2n,n) - n has the form x^m with m > 1.

			a(2) = 2 since C(2*1,1) + 1 = 3 is odd and C(2*2,2) + 2 = 8 is even.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..250 from Zhi-Wei Sun)

Cf. A000984, A232616, A232862, A232894.

L[m_,n_]:=Length[Union[Table[Mod[Binomial[2k,k]+k,n],{k,1,m}]]]
Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+3}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]

A232891 Least positive integer m <= n^2/2 + 3 such that {k*prime(k): k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 3, 4, 11, 7, 7, 10, 17, 43, 13, 51, 22, 51, 36, 31, 49, 64, 71, 119, 73, 86, 68, 141, 110, 153, 85, 83, 86, 144, 81, 174, 127, 115, 87, 122, 138, 143, 134, 133, 142, 211, 229, 152, 104, 109, 177, 259, 142, 194, 176, 196, 311, 312, 193, 243, 197, 396, 169, 156, 171

Offset: 1

Zhi-Wei Sun, Dec 02 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) For any positive integer n not equal to 3, the number n*prime(n) + 1 cannot be a power x^m with m > 1.
(iii) There are infinitely many positive integers n with n - 1, n + 1, n + prime(n), n + prime(n)^2, n^2 + prime(n), n^2 + prime(n)^2 all prime.

			a(3) = 4 since 1*prime(1) = 2, 2*prime(2) === 3*prime(3) == 0 (mod 3), and 4*prime(4) = 28 == 1 (mod 3).

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from Zhi-Wei Sun)

Cf. A000040, A232616, A232861, A232862.

L[m_,n_]:=Length[Union[Table[Mod[k*Prime[k],n],{k,1,m}]]]
Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+3}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]

cp's OEIS Frontend

A232398 Number of ways to write n = p + (2^k - k) + (2^m - m) with p prime and 0 < k <= m.

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A232862 Least positive integer m <= n^2/2 + 3 such that the set {prime(k) - k: k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

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A232894 Least positive integer m such that {Catalan(k) - k: k = 1, ..., m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

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A232898 Least positive integer m such that {C(2k,k) + k: k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

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A232891 Least positive integer m <= n^2/2 + 3 such that {k*prime(k): k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

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