A317474 The number of solutions to sigma(x) = sigma(x+1) below 10^n, where sigma(x) is the sum of divisors function (A000203).

0, 1, 3, 9, 24, 62, 113, 232, 533, 1097, 2295, 4804, 10135

Offset: 1

Author

Amiram Eldar, Jul 29 2018

Keywords

• nonn
• more

Comments

Data extracted from A002961.

The terms were calculated by:

a(1)-a(4) - Andrzej Makowski (1960);

a(5) - Mientka & Vogt (1970), Lal, Eldridge & Gillard (1972);

a(6)-a(7) - Hunsucker, Nebb & Stearns (1973);

a(8) - Pentti Haukkanen (1993);

a(9) - Jud McCranie (1997);

a(10) - T. D. Noe (2007);

a(11)-a(12) - T. D. Noe (2010);

a(13) - Giovanni Resta (2014).

It is not known whether there exist infinitely many natural numbers x for which sigma(x) = sigma(x+1) (see Sierpiński, p. 177). - Stefano Spezia, Jul 13 2025

Examples

			Below 10^3 there are 3 solutions x = 14, 206, 957, hence a(3) = 3.
		

References

• Wacław Sierpiński, Elementary Theory of Numbers, Warszawa, 1988.

Links

• Pentti Haukkanen, Some computational results concerning the divisor functions d(n) and sigma(n), The Mathematics Student, Vol. 62 (1993), pp. 166-168.
• John L. Hunsucker, Jack Nebb, and Robert E. Stearns, Jr., Computational results concerning some equations involving sigma(n), The Mathematics Student, Vol. 41 (1973), pp. 285-289; entire volume.
• M. Lal, C. Eldridge & P. Gillard, Solutions of sigma(n) = sigma(n+k), 1972, Review in Mathematics of Computation, Vol. 27, No. 123 (1973), p. 676.
• Andrzej Makowski, On Some Equations Involving Functions phi(n) and sigma(n), The American Mathematical Monthly, Vol. 67, No. 7 (1960), pp. 668-670; Correction, ibid., Volume 68, No. 7 (1961), p. 650.
• Walter E. Mientka and Richard L. Vogt, Computational results relating to problems concerning sigma(n), Matematicki Vesnik, Vol. 7, No. 51 (1970), pp. 35-36.

Crossrefs

Cf. A000203, A002961, A300285.

Programs

Mathematica

With[{s = Array[DivisorSigma[1,#]&, 10^5]}, Array[Count[Range[10^# - 1], ?(s[[#]] == s[[# + 1]] &)] &, IntegerLength@ Length@ s - 1]] (* after _Michael De Vlieger at A300285 *)

Formula

Conjecture: Limit_{n->oo} a(n)/A300285(n) = 1.