A254065 Vulgar fractions whose denominators are numbers ending with nine, the case 1/19.

1, 2, 4, 8, 6, 3, 7, 4, 9, 8, 7, 5, 1, 3, 6, 2, 5, 0, 1, 2, 4, 8, 6, 3, 7, 4, 9, 8, 7, 5, 1, 3, 6, 2, 5, 0, 1, 2, 4, 8, 6, 3, 7, 4, 9, 8, 7, 5, 1, 3, 6, 2, 5, 0, 1, 2, 4, 8, 6, 3, 7

Offset: 1

Author

Gary W. Adamson, Jan 24 2015

Keywords

• nonn
• easy
• base

Comments

The method provides an alternative for obtaining reversals of decimal expansions of 1/n, where n is of the form 10x + 9. We then access the reversal of the periodic sequence, noting that it matches the decimal expansion of 1/19 as shown in A021023: (.0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, ...). Page 23 of "Vedic Mathematics" states "In accordance with the sutra, we multiply it by 2" [since we have deleted the rightmost 9 and enhanced the remaining digit (a 1) by 1 = 2]. Then N. Ramamurthy states: "Similarly the multiplier for 49 is 5, for 149 is 15, for 12789 is 1279 and so on."

Examples

			Let 1/n = 1/19. Delete the 9 and increment the previous digit (1) by 1 to get 2, our multiplier M. Let the first term be 1, then continue multiplying by M, getting 1, 2, 4, 8, ... then the next term is 6 with a carryover of 1, giving (1, 2, 4, 8, 6, ...). The next term is 3 since 2 * 6 = 12; thus we mark down 3 (2 plus the carryover). The next term is 7 (being 2 * 3 plus the carryover of 1); proceeding in this way, we get 1, 2, 4, 8, 6, 3, 7, 4, 9, 8, 7, ...
		

References

• L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 172.
• N. Ramamurthy, "Vedic Mathematics, 30 Formulae Elucidated With Simple Examples", 9789382237273, published by N. Ramamurthy, http://ramamurthy.jaagruti.co.in, 2013, pp. 21-22.

Links

• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,-1,1).

Crossrefs

Cf. A021023.

Formula

Given 1/n, where n ends in 9, delete the 9 and use the previous set of digits (incremented by 1) as a multiplier M; let the first term = 1. Perform M * 1 and carry over the remainder if any, marking down the result. Perform M * result, adding the carryover for the next operation, and continue until the sequence repeats.

G.f.: -x*(1+x+2*x^2+4*x^3-2*x^4-3*x^5+4*x^6-3*x^7+5*x^8) / ( (x-1) *(1+x) *(x^2-x+1) *(x^6-x^3+1) ). - R. J. Mathar, May 24 2016

a(n) = (2^(n-1) mod 19) mod 10. - Ridouane Oudra, Jan 16 2023

Extensions

Edited by Jon E. Schoenfield, Jan 16 2023